HEREDITARY CONJUGACY SEPARABILITY OF RIGHT ANGLED … · class of special (or A-special, in the terminology of [34]) cube complexes, that admit a combinatorial local isometry to the

HEREDITARY CONJUGACY SEPARABILITY OF RIGHT ANGLEDARTIN GROUPS AND ITS APPLICATIONS

ASHOT MINASYAN

Abstract. We prove that finite index subgroups of right angled Artin groups are conju-gacy separable. We then apply this result to establish various properties of other classesof groups. In particular, we show that any word hyperbolic Coxeter group contains aconjugacy separable subgroup of finite index and has a residually finite outer automor-phism group. Another consequence of the main result is that Bestvina-Brady groups areconjugacy separable and have solvable conjugacy problem.

Contents

1. Introduction 1

2. Consequences of the main theorem 4

3. Hereditary conjugacy separability and Centralizer Conditions 8

4. Commuting retractions 11

5. Implications for the profinite topology 14

6. Some properties of right angled Artin groups 17

7. Special HNN-extensions 23

8. Proof of the main result 29

9. Applications to separability properties 39

10. Applications to outer automorphism groups 42

11. Applications to the conjugacy problem 43

12. Appendix: the Centralizer Condition in profinite terms 45

References 47

1. Introduction

If G is a group, the profinite topology PT (G) on G is the topology whose basic opensets are cosets to finite index normal subgroups in G. It follows that every finite index

2010 Mathematics Subject Classification. 20F36, 20E26, 20F55, 20F28, 20F10.Key words and phrases. Hereditary conjugacy separability, right angled Artin groups, graph groups,

partially commutative groups, Coxeter groups, Bestvina-Brady groups.1

2 ASHOT MINASYAN

subgroup K ≤ G is both closed and open in PT (G), and G, equipped with PT (G), isa topological group (that is, the group operations are continuous with respect to thistopology). This topology is Hausdorff if and only if the intersection of all finite indexnormal subgroups is trivial in G. In this case G is said to be residually finite.

We will say that a subset A ⊆ G is separable in G if A is closed in PT (G). Supposethat for every element g ∈ G, its conjugacy class gG := hgh−1 |h ∈ G ⊆ G is closed inPT (G). Then G is called conjugacy separable. In other words, G is conjugacy separableif and only if for any two non-conjugate elements x, y ∈ G there exists a homomorphismϕ from G to a finite group Q such that ϕ(x) is not conjugate to ϕ(y) in Q.

Conjugacy separability is evidently stronger than residual finiteness, and is (usually)much harder to establish. The following classes of groups are known to be conjugacyseparable: virtually free groups (J. Dyer [23]); virtually surface groups (A. Martino [39]);virtually polycyclic groups (V. Remeslennikov [51]; E. Formanek [26]); limit groups (S.Chagas and P. Zalesskii [12]) and, more generally, finitely presented residually free groups(S. Chagas and P. Zalesskii [11]).

Unfortunately, conjugacy separability does not behave very well under free construc-tions. V. Remeslennikov [52] and P. Stebe [57] showed that the free product of two con-jugacy separable groups is conjugacy separable. But so far we do not know of any globalcriteria which tell when an amalgamated product (or an HNN-extension) of conjugacyseparable groups is conjugacy separable. Perhaps the most general of local results can befound in [53], where L. Ribes, D. Segal and P. Zalesskii define a new class of conjugacyseparable groups X, which is closed under taking free products with amalgamation alongcyclic subgroups and contains all virtually free and virtually polycyclic groups. Note thatthere is no analogue of this result for HNN-extensions with associated cyclic subgroups,because an HNN-extension of the infinite cyclic group may fail to be residually finite, asit happens for many Baumslag-Solitar groups.

Let Γ be a finite simplicial graph, and let V and E be the sets of vertices and edges of Γrespectively. The right angled Artin group G, associated to Γ, is given by the presentation

(1.1) G := 〈V ‖uv = vu, whenever u, v ∈ V and (u, v) ∈ E〉.

In the literature, right angled Artin groups are also called graph groups or partiallycommutative groups. These groups received a lot of attention in the recent years: theyseem to be interesting from both combinatorial and geometric viewpoints (they are fun-damental groups of compact non-positively curved cube complexes). A good overview ofthe current results concerning right angled Artin groups can be found in R. Charney’spaper [13].

In the case when the finite graph Γ is a simplicial tree, conjugacy separability of theassociated right angled Artin group was proved by E. Green [28]. It also follows from theresult of Ribes, Segal and Zalesskii [53] mentioned above, because such tree groups areeasily seen to belong to the class X.

HEREDITARY CONJUGACY SEPARABILITY OF RIGHT ANGLED ARTIN GROUPS 3

Following [11], we will say that a group G is hereditarily conjugacy separable if everyfinite index subgroup of G is conjugacy separable.

Note that all of the classes of conjugacy separable groups that we mentioned above(possibly, with the exception of class X) consist, in fact, of hereditarily conjugacy separablegroups due to the obvious reason: these classes are closed under taking subgroups of finiteindex. However, there exist conjugacy separable, but not hereditarily separable groups.The first (infinitely generated) example, demonstrating this, was constructed by Chagasand Zalesskii in [11]. It is also possible to find finitely generated and finitely presentedexamples of this sort even among subgroups of right angled Artin groups (see [40]).

The main result of this work is the following theorem:

Theorem 1.1. Right angled Artin groups are hereditarily conjugacy separable.

Remark that a finite index subgroup of a right angled Artin group may not be aright angled Artin group itself. The following example was suggested to the author byM. Bridson:

Example 1.2. Let S be a finite group with a (finite) non-trivial second homology groupH2(S) (for instance, on can take S to be the alternating group A5, since H2(A5) ∼= Z/2Z).As we know, there is an epimorphism ψ : F → S for some finitely generated free groupF . Let K := (x, y) ∈ F × F |ψ(x) = ψ(y) be the fibre product associated to ψ.Observe that F × F is a right angled Artin group (associated to some finite completebipartite graph) and K is a finite index subgroup in it. By [8, Thm. A], H2(S) embedsinto H1(K) ∼= K/[K,K]. Therefore K is not isomorphic to any right angled Artin group,because the abelianization of a right angled Artin group is always a free abelian group,and, thus, it is torsion-free.

Generally speaking, we think that hereditary conjugacy separability is a lot strongerthan simply conjugacy separability. Corollaries in the next section can be viewed as aconfirmation of this.

Our proof of Theorem 1.1 is purely combinatorial and mostly self-contained (we usebasic properties of right angled Artin groups and HNN-extensions). The basic idea is toapproximate right angled Artin groups by HNN-extensions of finite groups (which are, ofcourse, virtually free). This is the main step of the proof. Once this is done, we can useknown properties of virtually free groups to obtain the desired results.

In Section 3 we study the Centralizer Condition, which, among other things, shows thata given conjugacy separable group is hereditarily conjugacy separable. This condition wasoriginally introduced by Chagas and Zalesskii in [11], but in a different form. In Sections4, 5 and 7 we develop machineries of commuting retractions and special HNN-extensionswhich are the two basic tools behind the proof of Theorem 1.1.

Acknowledgements. The author is very grateful to Frederic Haglund and Daniel Wisefor explaining their work in [34] and [33]. The author would also like to thank MartinBridson, Ilya Kazachkov, Graham Niblo and Pavel Zalesskii for discussions.

4 ASHOT MINASYAN

2. Consequences of the main theorem

Recall that a subgroup H of a group G is said to be a virtual retract of G, if there is afinite index subgroup K ≤ G such that H ≤ K and H is a retract of K (see Section 4 forthe definition).

It is not difficult to show (see Lemma 9.5) that a virtual retract of a hereditarily con-jugacy separable group is itself hereditarily conjugacy separable. Therefore, Theorem 1.1immediately yields

Corollary 2.1. If G is a right angled Artin group and H is a virtual retract of G, thenH is hereditarily conjugacy separable.

In view of the above corollary, it makes sense to define two classes of groups: the classVR will consist of all groups which are virtual retracts of finitely generated right angledArtin groups, and the class AVR will consist of groups that contain finite index subgroupsfrom the class VR.

Looking at the definition, it might seem that the class of right angled Artin groups isnot very large. However, the class of subgroups and virtual retracts of right angled Artingroups is quite rich and includes many interesting examples. For instance, in the famouspaper [5] M. Bestvina and N. Brady constructed subgroups of right angled Artin groupswhich have the property FP2 but are not finitely presented.

On the other hand, in the recent work [34] F. Haglund and D. Wise introduced a newclass of special (or A-special, in the terminology of [34]) cube complexes, that admit acombinatorial local isometry to the Salvetti cube complex (see [13]) of a right angled Artingroup (possibly, infinitely generated). They proved that the fundamental group of everyspecial complex X embeds into some right angled Artin group G (if X is not compactand has infinitely many hyperplanes, then the corresponding right angled Artin group Gwill be associated to an infinite graph Γ, and, hence, will not be finitely generated).

An important property established by Haglund and Wise in [34], states that if X isa compact A-special cube complex, then π1(X ) is a virtual retract of some right angledArtin group, i.e., π1(X ) ∈ VR. Therefore, using Corollary 2.1, we immediately obtain

Corollary 2.2. If H is the fundamental group of a compact A-special cube complex, thenH is hereditarily conjugacy separable.

Moreover, many other groups are virtually special, i.e., they possess finite index sub-groups that are fundamental groups of special cube complexes. Among virtually specialgroups are all Coxeter groups – see [33], fundamental groups of compact surfaces – see[17], fundamental groups of compact virtually clean square VH-complexes (introduced byWise in [58]) – see [34], graph braid groups (introduced by A. Abrams in [1]) – see [17],and some hyperbolic 3-manifold groups – see [15].

In this paper we will mainly discuss applications of Theorem 1.1 to Coxeter groups,even though similar corollaries can be derived for the other classes of virtually specialgroups listed above.


Recall that a Coxeter group is a group G given by the presentation

(2.1) G = 〈s1, . . . , sn ‖ (sisj)mij = 1, for all i, j with mij ∈ N〉,

where M := (mij) is a symmetric n × n matrix, whose entries satisfy the followingconditions: mii = 1 for every i = 1, . . . , n, mij ∈ N t ∞ and mij ≥ 2 whenever1 ≤ i < j ≤ n. In the case, when mij ∈ 2,∞ for all i 6= j, G is said to be a right angledCoxeter group.

For any Coxeter group G, G. Niblo and L. Reeves [47] constructed a locally finite, finitedimensional CAT(0) cube complex C on which G acts properly discontinuously. In [33]Haglund and Wise show that G has a finite index subgroup F such that F acts freely on Cand the quotient F \C is an A-special cube complex. In the case when G is right angled orword hyperbolic (in Gromov’s sense [29]), Niblo and Reeves proved that the action of Gon C is cocompact (see [47]). These results, combined with the virtual retraction theoremof Haglund and Wise mentioned above, imply that word hyperbolic (or right angled)Coxeter groups belong to the class AVR. Therefore, using Corollary 2.1 we achieve

Corollary 2.3. Every word hyperbolic (or right angled) Coxeter group G contains a finiteindex subgroup F which is hereditarily conjugacy separable.

Actually, as the paragraph above Corollary 2.3 shows, the conclusion of this corollaryholds for every finitely generated Coxeter group G, whose action on the correspondingNiblo-Reeves cube complex is cocompact. Such Coxeter groups were completely charac-terized by P.-E. Caprace and B. Muhlherr in [10].

The sole fact of existence of a conjugacy separable finite index subgroup F in G mayseem somewhat unsatisfactory. However, every Coxeter group is virtually torsion-free,and in a given Coxeter group G it is usually easy to find some torsion-free subgroup offinite index (for instance, if G is a right angled Coxeter group (2.1), then the kernel of thenatural homomorphism from G onto 〈s1〉2 × · · · × 〈sn〉2 ∼= (Z/2Z)n is torsion-free). Thefollowing statement is proved in Section 9:

Corollary 2.4. If G is a word hyperbolic Coxeter group, then every torsion-free finiteindex subgroup H ≤ G is hereditarily conjugacy separable.

The above corollary produces a lot of new examples of conjugacy separable groups.More generally, in Corollary 9.11 we show that every torsion-free word hyperbolic groupfrom the class AVR is hereditarily conjugacy separable.

Now, let us discuss some other consequences of the main result. Beside being a classicalsubject of group theory, conjugacy separability has two main applications. One of theapplications was found by E. Grossman in [30], where she showed that the outer auto-morphism group Out(G) of a finitely generated conjugacy separable group G is residuallyfinite, provided that every pointwise inner automorphism of G is inner (an automorphismφ ∈ Aut(G) is called pointwise inner if for every g ∈ G, φ(g) is conjugate to g in G).Thereafter, Grossman used this observation to prove that the mapping class group of acompact orientable surface is residually finite.

6 ASHOT MINASYAN

Note that for a finitely generated residually finite group G, the group of outer automor-phisms Out(G) need not be residually finite (this should be compared with the classicaltheorem of G. Baumslag [4] claiming that the automorphism group Aut(G) of a finitelygenerated residually finite group G is residually finite). This is a consequence of the resultof I. Bumagina and D. Wise ([9]) which asserts that for every finitely presented group Sthere exists a finitely generated residually finite group G such that Out(G) ∼= S.

In Section 6 we prove that pointwise inner automorphisms of right angled Artin groupsare inner (see Proposition 6.9). Thus Grossman’s result, combined with Theorem 1.1,gives

Theorem 2.5. For any right angled Artin group G, the group of outer automorphismsOut(G) is residually finite.

Presently not much is yet known about the outer automorphisms of an arbitrary rightangled Artin group G. M. Laurence [36] showed that Aut(G) (and, hence, Out(G)) isfinitely generated. More recently, M. Day [18] proved that Aut(G) (and, hence, Out(G))is finitely presented. In [14] R. Charney and K. Vogtmann showed that Out(G) is vir-tually torsion-free and has finite virtual cohomological dimension. Imposing additionalconditions on the finite graph Γ, corresponding to G, M. Gutierrez, A. Piggott and K.Ruane were able to extract more information about the structure of Aut(G) and Out(G)in [32].

On the other hand, in Section 10 we use a recent result of the author with D. Osin from[43] to prove the following theorem:

Theorem 2.6. If G ∈ AVR is a relatively hyperbolic group, then Out(G) is residuallyfinite.

Note that Theorem 2.5 is not a consequence of Theorem 2.6: it is not difficult to showthat a (non-cyclic) right angled Artin group G is relatively hyperbolic if and only if thegraph Γ, corresponding to G, is disconnected.

Applying Theorem 2.6 to our favorite class of groups from AVR, we achieve

Corollary 2.7. For any word hyperbolic Coxeter group G, Out(G) is residually finite.

Unlike automorphisms of right angled Artin groups, automorphism groups of Coxetergroups have already attracted a lot of attention. In many particular cases the structureof the (outer) automorphism group is known: see, for instance, P. Bahls’s paper [3] andreferences therein. However, because of its generality, the statement of Corollary 2.7seems to be new.

The second classical application of conjugacy separability was found by A. Mal’cev.As Mal’cev proved in [38] (see also [44]), a finitely presented conjugacy separable groupG has solvable conjugacy problem. Observe that finite presentability of G is importanthere, because the set of finite quotients of an infinitely presented group does not need tobe recursively enumerable.


It follows that the conjugacy problem is solvable for every group G ∈ VR: G is finitelypresented as a retract of a finitely presented group, and G is conjugacy separable byCorollary 2.1. However, most of the groups from the class VR that we discussed aboveare already known to have solvable conjugacy problem. Moreover, J. Crisp, E. Godelleand B. Wiest [16] showed that the conjugacy problem for fundamental groups of A-specialcube complexes can be solved in linear time.

Nevertheless, the property of hereditary conjugacy separability for a group G turns outto be powerful enough to yield conjugacy separability and solvability of the conjugacyproblem for many subgroups which are not virtual retracts of G – see Corollary 11.2.

Recall that a group G is called subgroup separable (or LERF) if every finitely generatedsubgroup H ≤ G is separable in G. In Section 11 we prove

Theorem 2.8. Let N be a normal subgroup of a right angled Artin group G such that thequotient G/N is subgroup separable. Then N is conjugacy separable. If, in addition, Nis finitely generated, then N has solvable conjugacy problem.

Note that requiring G/N to be subgroup separable cannot be dropped in the abovestatement: in [42] C. Miller gives an example of a finitely generated subgroup of F2 × F2

that has unsolvable conjugacy problem (here F2 denotes the free group of rank 2, and soF2 × F2 is the right angled Artin group associated to a square).

The second claim of Theorem 2.8 may seem surprising: in general we cannot useMal’cev’s result, mentioned above, to reach the needed conclusion, because the condi-tions imposed on N do not constrain it to be finitely presented. Indeed, let G be theright angled Artin group associated to a finite graph Γ and given by (1.1). Let NΓ be thekernel of the homomorphism ψ : G → Z satisfying ψ(v) = 1 for each v ∈ V , and let LΓ

be the simplicial flag complex, whose 1-skeleton is Γ.

J. Meier and L. VanWyk [41] proved that the group NΓ is finitely generated if andonly if the graph Γ is connected. And in [5] Bestvina and Brady showed that NΓ isfinitely presented if and only if the complex LΓ is simply connected. In the case when Γis connected, we will say that NΓ is the Bestvina-Brady group associated to Γ.

For example, if the graph Γ is a cycle of length at least 4, then NΓ is finitely generated,but not finitely presented. Obviously, the quotient G/NΓ

∼= Z is subgroup separable,hence, by Theorem 2.8, NΓ is conjugacy separable and has solvable conjugacy problem.More generally, we have the following corollary:

Corollary 2.9. If N is a finitely generated normal subgroup of a right angled Artin groupG such that G/N is abelian, then N is hereditarily conjugacy separable and has solv-able conjugacy problem. In particular, Bestvina-Brady groups are hereditarily conjugacyseparable and have solvable conjugacy problem.

Corollary 2.9 is a direct consequence of Corollary 11.3 (proved at the end of Section 11),that covers the more general case when G/N is polycyclic. We have chosen to mentionthe particular situation when the quotient G/N abelian in Corollary 2.9, because in thiscase one can tell whether or not the given normal subgroup N is finitely generated, using

8 ASHOT MINASYAN

Bieri-Neumann-Strebel invariants, which were studied for right angled Artin groups byMeier and VanWyk in [41].

After finishing this paper the author learned that the positive solution of the conjugacyproblem for the group N from Theorem 2.8 (or Corollary 2.9) has been known before.This follows from a more general result of M. Bridson [7], claiming that a normal subgroupN of a bicombable group G has solvable conjugacy problem, provided G/N has solvablegeneralized word problem (see [7] for the definitions). Indeed, any right angled Artingroup G acts properly and cocompactly on a CAT(0) space (which is the universal coverof the corresponding compact non-positively curved Salvetti cube complex), therefore Gis bicombable by a theorem of J. Alonso and M. Bridson [2]. And if N CG, the subgroupseparability of G/N implies that G/N has solvable generalized word problem, by Mal’cev’sresult [38].

Nevertheless, the statement claiming that N is conjugacy separable in Theorem 2.8(resp. hereditarily conjugacy separable in Corollary 2.9) is new. Our solution of theconjugacy problem for N uses a Mal’cev-type argument and can be viewed as anotherapplication of hereditary conjugacy separability of G.

3. Hereditary conjugacy separability and Centralizer Conditions

First, let us specify some notations. If G is a group, H ≤ G is a subgroup and g ∈ G,then the H-conjugacy class of the element g ∈ G is the subset gH := hgh−1 |h ∈ H ⊆ G.For any A ⊆ G, we denote AH := hah−1 | a ∈ A, h ∈ H. The H-centralizer of g ∈ G isthe subgroup CH(g) := h ∈ H | hg = gh ≤ G. For an epimorphism ψ : G→ F from Gonto a group F , ψ−1 : 2F → 2G will denote the corresponding full preimage map. If A,Bare two subsets of G then their product AB is defined as the subset ab | a ∈ A, b ∈ B ⊆G. Note that if either A or B is empty, then the product AB is empty as well.

Definition 3.1. Suppose that G is a group. We will say that G satisfies the CentralizerCondition (briefly, CC), if for every finite index normal subgroup KCG and every g ∈ Gthere is a finite index normal subgroup LCG such that L ≤ K and

(3.1) CG/L(g) ⊆ ψ(CG(g)K

)in G/L

(where ψ : G→ G/L is the natural epimorphism and g := ψ(g)).

Note that (3.1) is equivalent to ψ−1(CG/L(g)

)⊆ CG(g)K in G, since ker(ψ) = L ≤ K.

The idea behind this condition is to provide control over the growth of centralizers infinite quotients of G. If the group G is residually finite, the Centralizer Condition CC can

be reformulated in terms of the topology on the profinite completion G of the group G.In the Appendix to this paper (see Corollary 12.2) we prove that the condition CC fromDefinition 3.1 is equivalent to the following:

(3.2) CG(g) = CG(g) in G, for every g ∈ G

(where CG(g) denotes the closure of CG(g) in G).


Originally the condition (3.2) appeared in the recent work of Chagas and Zalesskii [11],where they proved that a conjugacy separable group G satisfying (3.2) is hereditarilyconjugacy separable (see [11, Prop. 3.1]). We will actually show that, provided G isconjugacy separable, this condition is equivalent to hereditary conjugacy separability:

Proposition 3.2. Let G be a group. Then the following are equivalent:

(a) G is hereditarily conjugacy separable;(b) G is conjugacy separable and satisfies CC.

Before proving Proposition 3.2, let us define two more conditions.

Definition 3.3. Let G be a group, H ≤ G and g ∈ G. We will say that the pair(H, g) satisfies the Centralizer Condition in G (briefly, CCG), if for every finite indexnormal subgroup K CG there is a finite index normal subgroup LCG such that L ≤ Kand CH(g) ⊆ ψ (CH(g)K) in G/L, where ψ : G → G/L is the natural homomorphism,H := ψ(H) ≤ G/L, g := ψ(g) ∈ G/L.

The subgroup H will be said to satisfy the Centralizer Condition in G (briefly, CCG)if for each g ∈ G, the pair (H, g) has CCG.

Now, let demonstrate why the Centralizer Conditions are useful.

Lemma 3.4. Suppose that G is a group, H ≤ G and g ∈ G. Assume that the pair (G, g)satisfies CCG and the conjugacy class gG is separable in G. If the double coset CG(g)His separable in G, then the H-conjugacy class gH is also separable in G.

Proof. Consider any element y ∈ G with y /∈ gH .

If y /∈ gG, then, using the separability of gG, we can find a finite quotient Q of G anda homomorphism φ : G → Q so that φ(y) /∈ φ(g)Q. Hence φ(y) /∈ φ(g)φ(H) = φ(gH), asrequired.

Therefore we can assume that y = zgz−1 for some z ∈ G. If there existed an element f ∈CG(g)∩z−1H, then zf ∈ H and y = zgz−1 = (zf)g(zf)−1 ∈ gH , leading to a contradictionwith our assumption on y. Hence CG(g) ∩ z−1H = ∅, i.e., z−1 /∈ CG(g)H. Since CG(g)His separable in G, there is K CG such that |G : K| <∞ and z−1 /∈ CG(g)HK. Now, thecondition CCG implies that there exists a finite index normal subgroup LCG such thatL ≤ K and CG/L(ψ(g)) ⊆ ψ(CG(g)K), where ψ : G→ G/L is the natural epimorphism.

We claim that ψ(y) /∈ ψ(gH) in G/L. Indeed, if ψ(y) = ψ(hgh−1) for some h ∈ H,then ψ(z−1h) ∈ CG/L(ψ(g)). Hence ψ(z−1) ∈ CG/L(ψ(g))ψ(H) ⊆ ψ(CG(g)KH), i.e.,z−1 ∈ CG(g)KHL = CG(g)HK because L ≤ K CG. But this yields a contradiction withthe construction of K.

Therefore we have found an epimorphism ψ from G to a finite group G/L such thatthe image of y does not belong to the image of gH . Hence gH is separable in G.

Observe that for a subgroup H of a group G and any subset S ⊆ H, if S is closedin PT (G), then S is closed in PT (H). Therefore Lemma 3.4 immediately implies thefollowing:

10 ASHOT MINASYAN

Corollary 3.5. Let G be a conjugacy separable group satisfying CC, and let H ≤ G bea subgroup such that CG(h)H is separable in G for every h ∈ H. Then H is conjugacyseparable. Moreover, for each h ∈ H the H-conjugacy class hH is closed in PT (G).

It is not difficult to see that Lemma 3.4 has a partial converse (we leave its proof as anexercise for the reader):

Remark 3.6. Assume that H is a subgroup of a group G and g ∈ G is an arbitrary element.If gH is separable in G then the double coset CG(g)H is separable in G.

In this paper we are going to use a different converse to Lemma 3.4:

Lemma 3.7. Let G be a group. Suppose that H ≤ G, g ∈ G, K C G and |G : K| < ∞.If the subset gH∩K is separable in G, then there is a finite index normal subgroup LC Gsuch that L ≤ K and CH(g) ⊆ ψ (CH(g)K) in G/L (in the notations of Definition 3.3).

Proof. Denote k := |H : (H ∩ K)| ≤ |G : K| < ∞. Then H =⊔ki=1 zi(H ∩ K) for

some z1, . . . , zk ∈ H. Renumbering the elements zi, if necessary, we can suppose thatthere is l ∈ 0, 1, . . . , k such that whenever 1 ≤ i ≤ l, z−1

i gzi /∈ gH∩K , and wheneverl + 1 ≤ j ≤ k, z−1

j gzj ∈ gH∩K in G.

By the assumptions, there exists a finite index normal subgroup L C G such thatz−1i gzi /∈ gH∩KL whenever 1 ≤ i ≤ l. Moreover, after replacing L with L ∩ K, we can

assume that L ≤ K.

Let ψ be the natural epimorphism from G to G/L and consider any element x ∈ CH(g).Then x = ψ(x) for some x ∈ H, and ψ(x−1gx) = ψ(g) in G/L, i.e., x−1gx ∈ gL in G.As we know, there is i ∈ 1, . . . , k and y ∈ H ∩K such that x = ziy. Consequently,z−1i gzi ∈ ygLy−1 = ygy−1L ⊆ gH∩KL. Hence, i ≥ l + 1, that is, z−1

i gzi = ugu−1 for someu ∈ H ∩K.

Thus ziu ∈ CH(g) and x = ziy = (ziu)(u−1y) ∈ CH(g)(H ∩K) ⊆ CH(g)K. Thereforewe proved that x ∈ ψ(CH(g)K) in G/L for every x ∈ CH(g). This yields the inclusionCH(g) ⊆ ψ (CH(g)K) in G/L, as required.

We are now ready to prove Proposition 3.2.

Proof of Proposition 3.2. First let us assume (b). Consider an arbitrary finite index sub-group H ≤ G. For every h ∈ H the double coset CG(h)H is a finite union of left cosetsmodulo H, hence it is separable in G. Therefore, by Corollary 3.5, H is conjugacy sepa-rable. That is, (b) implies (a).

Now, assume that G is hereditarily conjugacy separable. We need to show that Gsatisfies CC. Take any g ∈ G and any K C G with |G : K| < ∞. Observe that thesubgroup H := K〈g〉 ≤ G has finite index in G, and gH = gK = gH∩K . Since H isconjugacy separable, gH is closed in PT (H), but then it is also closed in PT (G) becauseany finite index subgroup of H has finite index in G. Therefore gH∩K = gH is separablein G, and so we can apply Lemma 3.7 to find the finite index normal subgroup L C Gfrom its claim. Hence the group G satisfies CC.


4. Commuting retractions

In this section we establish certain properties of commuting retractions that constitutethe core of our approach to studying residual properties of right angled Artin groups. Thisapproach is based on a simple observation that canonical retractions of a right angled Artingroup onto its special subgroups pairwise commute (see Remark 6.1 in Section 6).

Let G be a group and let H be a subgroup of G. Recall that an endomorphism ρH :G→ G is called a retraction of G onto H if ρH(G) = H and ρH(h) = h for every h ∈ H.In this case H is said to be a retract of G. Note that ρH ρH = ρH . If H is a retract andg ∈ G, then the subgroup F := gHg−1 ≤ G, conjugate to H in G, is also a retract. Thecorresponding retraction ρF ∈ End(G) is given by the formula ρF (x) := gρH(g−1xg)g−1

for all x ∈ G.

The following observation is very useful:

Lemma 4.1. Let H be a retract of a group G and let ρH : G → G be the correspondingretraction. Suppose that MCG satisfies ρH(M) ⊆M . Then the retraction ρH canonicallyinduces a retraction ρH : G/M → G/M of G/M onto the natural image H of H in G/M ,defined by the formula ρH(gM) = ρH(g)M for all gM ∈ G/M .

Proof. Evidently, it is enough to check that ρH is well-defined. If g1M = g2M for someg1, g2 ∈ G, then f = g−1

2 g1 ∈M , g1 = g2f and ρH(f) ∈M . Hence

ρH(g1M) = ρH(g1)M = ρH(g2)ρH(f)M = ρH(g2)M = ρH(g2M),

as required.

Assume that H and F are two retracts of a group G and ρH , ρF ∈ End(G) are thecorresponding retractions. We will say ρH commutes with ρF if they commute as elementsof the monoid of endomorphisms End(G), i.e., if ρH(ρF (g)) = ρF (ρH(g)) for all g ∈ G.

Remark 4.2. If the retractions ρH and ρF commute then ρH(F ) = H ∩ F = ρF (H) andthe endomorphism ρH∩F := ρH ρF = ρF ρH is a retraction of G onto H ∩ F .

Indeed, obviously the restriction of ρH∩F to H ∩ F is the identity map. And ρH∩F (G) ⊆ρH(G) ∩ ρF (G) = H ∩ F , hence ρH∩F (G) = H ∩ F . Consequently ρH(F ) = ρH(ρF (G)) =ρH∩F (G) = H ∩ F . Similarly, ρF (H) = H ∩ F .

In the next proposition we establish an important property of commuting retractionsthat could be of independent interest.

Proposition 4.3. Let H1, . . . , Hm be retracts of a group G such that the correspondingretractions ρH1 , . . . , ρHm pairwise commute. Then for any finite index normal subgroupKCG there is a finite index normal subgroup M CG such that M ≤ K and ρHi

(M) ⊆Mfor each i = 1, . . . ,m. Consequently, for every i = 1, . . . ,m, the retraction ρHi

canonicallyinduces a retraction ρHi

of G/M onto the image Hi of Hi in G/M .

Proof. The second claim of the proposition follows from Lemma 4.1, so it suffices toconstruct the subgroup M CG with the needed properties.

12 ASHOT MINASYAN

If J = i1, . . . , ik is a subset of the finite set I := 1, 2, . . . ,m, we define the retractionρJ of G onto

⋂j∈J Hj by

ρJ := ρHi1 ρHi2

· · · ρHik.

This makes sense since our retractions pairwise commute. When J = ∅, ρJ will be theidentity map of G.

Now, for every subset J of I = 1, 2, . . . ,m we define the subgroup DJ ≤ G as follows.First we set DI :=

⋂mi=1Hi∩K – a finite index normal subgroup of (H1∩· · ·∩Hm). Next,

if J is a proper subset of I, we define DJ recursively, according to the following formula:

(4.1) DJ := ρJ

⋂i∈I\J

ρ−1J∪i(DJ∪i)

∩K,where ρ−1

J∪i(DJ∪i) denotes the full preimage (under ρJ∪i) of DJ∪i in G.

Since the intersection of a finite number of finite index normal subgroups is again afinite index normal subgroup, and images, as well as full preimages, of finite index normalsubgroups under homomorphisms are again normal and of finite index (in their respectivegroups), we see that DJ is normal and has finite index in ρJ(G) =

⋂j∈J Hj. Thus, if we

set M := D∅ =⋂i∈I ρ

−1Hi

(Di) ∩K, we shall have M CG, |G : M | <∞ and M ≤ K.

If J ⊂ I and i ∈ I \ J , using (4.1) and the fact that ρi ρJ = ρJ∪i, we can observethat ρi(DJ) ⊆ DJ∪i.

On the other hand, let us show that DJ∪i ⊆ DJ . We will use induction on thecardinality |I \J |. If |I \J | = 1 then I = J ti. And if g ∈ DJ∪i = DI =

⋂i∈I Hi∩K,

then ρI(g) = g, therefore g ∈ ρ−1I (DI) and g = ρJ(g) ∈ ρJ

(ρ−1I (DI)

). Thus g ∈ DJ .

Suppose, now, that the statement has been proved for all proper subsets J ′ of I with|J ′| > |J |. Take any i ∈ I\J and consider an element g ∈ DJ∪i ≤

⋂j∈J∪iHj∩K. Then

ρJ∪i(g) = g, therefore g ∈ ρ−1J∪i(DJ∪i). We need to show that for any i′ ∈ I \(J∪i),

g ∈ ρ−1J∪i′(DJ∪i′), or, equivalently, that ρJ∪i′(g) ∈ DJ∪i′. But

ρJ∪i′(g) = ρi′(ρJ(g)) = ρi′(g) ∈ ρi′(DJ∪i) ⊆ DJ∪i,i′.

And, since DJ∪i,i′ ⊆ DJ∪i′ by the induction hypothesis, we can conclude that g ∈⋂i′∈I\J ρ

−1J∪i′(DJ∪i′). Recalling that g ∈

⋂j∈J Hj ∩ K, we achieve g = ρJ(g) ∈ DJ .

Thus DJ∪i ⊆ DJ and the inductive step is established.

We are now able to show that ρHi(M) ⊆ M for every i ∈ I. Indeed, since ρHi

(M) ⊆Di, it is enough to check that Di ⊆ M . Take any j ∈ I. As we proved, ρHj

(Di) =

ρj(Di) ⊆ Di∪j ⊆ Dj. Therefore, Di ⊆ ρ−1Hj

(Dj) for each j ∈ J . By definition,Di ≤ K, consequently, for any i ∈ I, we achieve

ρHi(M) ⊆ Di ⊆

⋂i∈I

ρ−1Hi

(Di) ∩K = M,

as required.


The next observation is an easy consequence of the definition of M using the formula(4.1).

Remark 4.4. In Proposition 4.3, if G/K is a finite p-group for some prime number p, thenso is G/M .

Given two subgroups H and F of a group G, it is usually difficult to find quotient-groups Q of G such that the image of the intersection of H and F in Q coincides with theintersection of the images of these subgroups in Q. However, in the case when H and Fare retracts and the corresponding retractions commute this will be automatic for manyquotients of G.

Lemma 4.5. Suppose that the retractions ρH , ρF ∈ End(G) commute, and M C G is anormal subgroup satisfying ρH(M) ⊆M and ρF (M) ⊆M . Then ϕ(H∩F ) = ϕ(H)∩ϕ(F )in G/M , where ϕ : G→ G/M is the natural epimorphism.

Proof. By Lemma 4.1 ρH and ρF canonically induce retractions ρϕ(H) and ρϕ(F ) of G/Monto ϕ(H) and ϕ(F ) respectively.

Clearly, ϕ(H∩F ) ⊆ ϕ(H)∩ϕ(F ), and so, we only need to establish the inverse inclusion.Consider an arbitrary g ∈ ϕ(H)∩ϕ(F ). Then g = ϕ(g) for some g ∈ G, and ρϕ(F )(g) = g,ρϕ(H)(g) = g. Therefore

g = ρϕ(H)

(ρϕ(F )(ϕ(g))

)= ρϕ(H) (ϕ(ρF (g))) = ϕ (ρH(ρF (g))) ∈ ϕ(H ∩ F ),

where the last inclusion follows from Remark 4.2. Thus ϕ(H) ∩ ϕ(F ) ⊆ ϕ(H ∩ F ).

Lemma 4.5 allows to obtain the first interesting application of Proposition 4.3.

Corollary 4.6. Let H1, . . . , Hm be retracts of a group G such that the correspondingretractions ρH1 , . . . , ρHm pairwise commute. Then for any finite index normal subgroupKCG there is a finite index normal subgroup M CG such that M ≤ K and ρHi

(M) ⊆Mfor each i = 1, . . . ,m. Moreover, if ϕ : G→ G/M denotes the natural epimorphism, thenϕ(⋂mi=1 Hi) =

⋂mi=1 ϕ(Hi).

Proof. First we apply Proposition 4.3 to find the finite index normal subgroup M fromits claim. The last statement of the corollary will be proved by induction on m. Ifm = 1 there is nothing to prove. So let us assume that m ≥ 2 and we have alreadyshown that ϕ(

⋂m−1i=1 Hi) =

⋂m−1i=1 ϕ(Hi). Using Remark 4.2 we see that the map ρF :=

ρH1 · · · ρHm−1 ∈ End(G) is a retraction of G onto F :=⋂m−1i=1 Hi. By Proposition 4.3,

ρHi(M) ⊆M for each i = 1, . . . ,m, therefore

ρF (M) = (ρH1 · · · ρHm−2)(ρHm−1(M)

)⊆

(ρH1 · · · ρHm−3)(ρHm−2(M)

)⊆ · · · ⊆ ρH1(M) ⊆M.

By the assumptions, the retractions ρF and ρHm commute, hence we can applyLemma 4.5 to conclude that ϕ(F ∩ Hm) = ϕ(F ) ∩ ϕ(Hm). But ϕ(F ) =

⋂m−1i=1 ϕ(Hi)

by the induction hypothesis, consequently ϕ(⋂mi=1Hi) = ϕ(F ∩ Hm) =

⋂mi=1 ϕ(Hi), and

the proof is finished.

14 ASHOT MINASYAN

Let us now give an example which shows that the statements of Corollary 4.6 andLemma 4.5 are no longer true if the retractions do not commute.

Example 4.7. Let S be any infinite simple group, and let H be an arbitrary group pos-sessing non-trivial finite quotients. Set G := H ∗S, fix an element s ∈ S \ 1 and denoteF := sHs−1 ≤ G. Evidently H is a retract of G, where the retraction ρH : G → G of Gonto H is the identity on H and trivial on S. Clearly the endomorphism ρF ∈ End(G)defined by ρF (g) := sρH(s−1gs)s−1 for every g ∈ G, is a retraction of G onto F .

It is not difficult to see that the retractions ρH and ρF do not commute (for instance,because (ρH ρF )(G) = H, (ρF ρH)(G) = F and H ∩ F = 1).

If K CG is an arbitrary proper normal subgroup of finite index, then S ⊂ K (becauseS has no non-trivial finite quotients), hence the kernel ker(ρH) (which is equal to thenormal closure of S in G) is contained in K. Consequently, ρH(K) ⊆ ρ−1

H (ρH(K)) ⊆ K.Similarly, ρF (K) ⊆ K.

Observe that H ∩ F = 1 by construction. Denote by Q the quotient G/K and letϕ : G→ Q be the natural epimorphism. Since s ∈ S ≤ ker(ϕ) we see that

ϕ(H) ∩ ϕ(F ) = ϕ(H) = Q 6= 1 = ϕ(H ∩ F ).

That is, in any non-trivial finite quotient Q of G the intersection of the images of H andF is strictly larger than the image of H ∩ F .

5. Implications for the profinite topology

Throughout this section we will assume that A and B are retracts of a group G suchthat the corresponding retractions ρA ∈ End(G) and ρB ∈ End(G) commute. Our goalhere is to establish several consequences of these settings for the profinite topology on G.

Lemma 5.1. For arbitrary elements x, y ∈ G define α := ρA (ρB(x)x−1)xρB (x−1) ∈AxB ⊆ G and β := ρA (ρB(y)y−1) yρB (y−1) ∈ AyB ⊆ G. Then the following twoconditions are equivalent:

(i) y ∈ AxB;(ii) β ∈ αA∩B.

Proof. Observe that y ∈ AxB if and only if AyB = AxB, which is equivalent to AβB =AαB. Thus y ∈ AxB if and only if β ∈ AαB.

To show that (i) implies (ii), suppose that there are a ∈ A and b ∈ B such that

(5.1) β = aαb.

By definition, ρA(α) = 1 = ρA(β), hence 1 = ρA(a)ρA(b). Therefore, (5.1) implies thata = ρA(a) = ρA(b−1) ∈ ρA(B) = A ∩B (by Remark 4.2).

Now, since ρB ρA = ρA ρB we have ρB(α) = 1 = ρB(β). Therefore, applying ρB toboth sides of the equality (5.1), we get 1 = ρB(a)ρB(b) = ab because a, b ∈ B. Henceb = a−1 ∈ A ∩B and β = aαa−1 ∈ αA∩B.


Now, suppose that β ∈ αA∩B. Then β ∈ (A ∩ B)α(A ∩ B) ⊆ AαB. Thus (ii) implies(i).

Let us look at the proof of the above lemma in the particular case when y = x. Thenwe see that β = α, and

A ∩ xBx−1 = γ−1(A ∩ αBα−1

)γ, where γ := ρA

(ρB(x)x−1

)∈ A.

We also see that a ∈ A ∩ αBα−1 if and only if there is b ∈ B such that α = a−1αb.But, as we showed in the proof of Lemma 5.1, this can happen only if b = a ∈ A ∩ B.I.e, αa = aα and a ∈ A∩B, which is equivalent to a ∈ CA∩B(α). Thus, in this particularcase we obtain the following statement:

Lemma 5.2. If x ∈ G is an arbitrary element, then

A ∩ xBx−1 = γ−1CA∩B(α)γ in G,

where α := ρA (ρB(x)x−1)xρB (x−1) ∈ AxB and γ := ρA (ρB(x)x−1) ∈ A .

Combining Lemma 5.1 with Corollary 4.6 we achieve

Lemma 5.3. Consider any x ∈ G and denote α := ρA (ρB(x)x−1)xρB (x−1) ∈ AxB ⊆ G.If the conjugacy class αA∩B is separable in G, then the double coset AxB is also separablein G.

Proof. Suppose that an element y ∈ G satisfies y /∈ AxB. By Lemma 5.1, this is equivalentto β /∈ αA∩B, where β := ρA (ρB(y)y−1) yρB (y−1). Since αA∩B is separable, there is afinite index normal subgroup K C G such that ψ(β) /∈ ψ

(αA∩B

)= ψ(α)ψ(A∩B), where

ψ : G→ G/K is the canonical epimorphism.

By Corollary 4.6, there exists a finite index normal subgroup M C G such that M ≤K, ρA(M) ⊆ M , ρB(M) ⊆ M and ϕ(A ∩ B) = ϕ(A) ∩ ϕ(B), where ϕ is the naturalepimorphism from G to G/M . Since ψ factors through ϕ, we can conclude that ϕ(β) /∈ϕ(α)ϕ(A∩B) = ϕ(α)ϕ(A)∩ϕ(B). But by Lemma 4.1, there are canonically induced commutingretractions ρA and ρB of G/M onto A := ϕ(A) and B := ϕ(B) respectively. Moreover,letting x := ϕ(x), y := ϕ(y) and using the definition of the retractions ρA and ρB, weobtain ϕ(α) = ρA (ρB(x)x−1) xρB (x−1) and ϕ(β) = ρA (ρB(y)y−1) yρB (y−1). Therefore,by Lemma 5.1, applied to the retracts A and B in G/M , we have y /∈ AxB. That is,ϕ(y) /∈ ϕ(AxB). Hence the double coset AxB is separable in G.

Since the 1A∩B = 1 is separable in G whenever G is residually finite, we have thefollowing immediate consequence of Lemma 5.3.

Corollary 5.4. If A and B are retracts of a residually finite group G such that thecorresponding retractions commute, then the double coset AB is separable in G.

The statement of Corollary 5.4 has been known before – see, for example, [34, Lemma9.3], but the proof that we have presented here is new.

The following statement is well-known:

16 ASHOT MINASYAN

Lemma 5.5. Suppose that G is a residually finite group and A ≤ G is a retract of G. Ifa subset S ⊆ A is closed in PT (A), then S is closed in PT (G).

Proof. We will show that S coincides with its closure clG(S) in the profinite topology onG. By Corollary 5.4 the subgroup A = AA is closed in PT (G), hence clG(S) ⊆ A. Now, ifa ∈ A\S, then there is a homomorphism φ : A→ Q from A to a finite group Q such thatφ(a) /∈ φ(S). Since A is a retract of G, we have a homomorphism ψ : G→ Q defined byψ := φ ρA. Evidently ψ(a) = φ(a) /∈ φ(S) = ψ(S), hence a /∈ clG(S). Thus S = clG(S),as required.

Now, the reason why we need Lemma 5.2, is because it tells us that if one can controlthe A ∩ B-centralizers in G, then one can also control the intersections of conjugates ofthe retracts A and B. As it can be seen from Example 4.7, in general we may not be ableto find a finite quotient Q of G, in which the image of the intersection of two particularconjugates of A and B is equal to the intersection of their images. However, providedthat a certain Centralizer Condition is satisfied, we can find many finite quotients Q ofG where these two sets are very close to each other.

Lemma 5.6. Let x be an element of G and let α := ρA (ρB(x)x−1)xρB (x−1) ∈ G.Suppose that the pair (A ∩ B,α) satisfies the Centralizer Condition in G. Then for anyfinite index normal subgroup K C G there exists a finite index normal subgroup M C Gsuch that M ≤ K, ρA(M) ⊆M , ρB(M) ⊆M and ϕ(A)∩ϕ(xBx−1) ⊆ ϕ(A∩xBx−1)ϕ(K)in G/M , where ϕ : G→ G/M is the natural epimorphism.

Proof. By Lemma 5.2, A ∩ xBx−1 = γ−1CA∩B(α)γ, where γ := ρA (ρB(x)x−1) ∈ A.Since the pair (A ∩ B,α) has CCG, there is a subgroup L C G of finite index in G, suchthat L ≤ K and ψ−1

(Cψ(A∩B)(ψ(α))

)⊆ CA∩B(α)K in G, where ψ : G → G/L is the

natural epimorphism. Applying Corollary 4.6 to A, B and L we find a finite index normalsubgroup M CG, together with the epimorphism ϕ : G→ G/M , such that M ≤ L ≤ K,ρA(M) ⊆M , ρB(M) ⊆M and ϕ(A) ∩ ϕ(B) = ϕ(A ∩B).

By Lemma 4.1, ρA and ρB canonically induce retractions ρA and ρB of G/M ontoA := ϕ(A) and B := ϕ(B) respectively. Obviously ρA commutes with ρB in End(G/M),because ρA commutes with ρB in End(G).

Denote x := ϕ(x), α = ρA (ρB(x)x−1) xρB (x−1) ∈ G/M and γ := ρA (ρB(x)x−1) ∈ A.Observe that α = ϕ(α) and γ = ϕ(γ) by the definitions of ρA and ρB. Then by Lemma 5.2,A∩ xBx−1 = γ−1CA∩B(α)γ in G/M . Therefore, recalling that A∩ B = ϕ(A∩B), we get

ϕ−1(A ∩ xBx−1

)= ϕ−1

(γ−1CA∩B(α)γ

)= γ−1ϕ−1

(Cϕ(A∩B)(α)

)γ.

But since ψ factors through ϕ (as ker(ϕ) = M ≤ L = ker(ψ)), we obviously have

ϕ−1(Cϕ(A∩B)(α)

)⊆ ψ−1

(Cψ(A∩B)(ψ(α))

)⊆ CA∩B(α)K.

Hence we can conclude that ϕ−1(A ∩ xBx−1

)⊆ γ−1CA∩B(α)γK = (A ∩ xBx−1)K. Con-

sequently, ϕ(A) ∩ ϕ(xBx−1) = A ∩ xBx−1 ⊆ ϕ(A ∩ xBx−1)ϕ(K), and the lemma isproved.


In this paper we will need one more criterion for separability of specific double cosetsin G. In a certain sense it generalizes Remark 3.6.

Lemma 5.7. Consider arbitrary elements x, g ∈ G. Denote D := xBx−1 ≤ G andα := ρA (ρB(x)x−1)xρB (x−1) ∈ G. Suppose that the conjugacy classes αA∩B and gA∩D

are separable in G, and the pair (A∩B,α) satisfies CCG. Then the double coset CA(g)Dis separable in G.

Proof. Consider any z ∈ G with z /∈ CA(g)D. First, suppose that z /∈ AD. Since αA∩B

is separable in G, Lemma 5.3 implies that AxB is separable, hence AD = (AxB)x−1

is separable as well (because multiplication by a fixed group element on the right isa homeomorphisms of G with respect to the profinite topology). Therefore there is afinite index normal subgroup N C G such that z /∈ ADN , hence z /∈ CA(g)DN becauseCA(g) ≤ A.

Thus we can assume that z ∈ AD, i.e., there exist a0 ∈ A and d0 ∈ D such thatz = a0d0. Since z /∈ CA(g)D, y := zd−1

0 /∈ CA(g)(A ∩ D). Consequently, for everyh ∈ A ∩D, (yh)g(yh)−1 6= g, i.e., y−1gy 6= hgh−1, implying that y−1gy /∈ gA∩D in G.

Now, the separability of gA∩D in G implies that there is a finite index normal subgroupK C G such that y−1gy /∈ gA∩DK. And, by Lemma 5.6, we can find a finite indexnormal subgroup M C G such that M ≤ K and ϕ(A) ∩ ϕ(D) ⊆ ϕ(A ∩ D)ϕ(K), whereϕ : G→ G/M is the natural epimorphism. Let K, A, D, y and g denote the ϕ-images ofK, A, D, y and g respectively.

Since K C G/M , we have gA∩D ⊆ gϕ(A∩D)K ⊆ gϕ(A∩D)K and y−1gy /∈ gϕ(A∩D)K asM ≤ K. Hence y−1gy /∈ gA∩D.

To finish the proof, it remains to show that ϕ(z) /∈ ϕ(CA(g)D). Suppose, on thecontrary, that there exist a ∈ CA(g) and d ∈ D such that ϕ(z) = ϕ(ad). Then ϕ(a0d0) =ϕ(ad), thus h := ϕ(a−1a0) = ϕ(dd−1

0 ) ∈ A ∩ D, and ϕ(z) = ϕ(a)hϕ(d0). Consequently,y = ϕ(z)ϕ(d−1

0 ) = ϕ(a)h and

y−1gy = h−1ϕ(a−1ga)h = h−1gh ∈ gA∩D,contradicting to our construction.

Thus, for every z /∈ CA(g)D we found M C G with |G : M | < ∞ such that z /∈CA(g)DM . Therefore the double coset CA(g)D is separable in G.

6. Some properties of right angled Artin groups

In this section we recall a few properties of right angled Artin groups, which will be usedin the proof of the main result. At the end of the section we prove that every pointwiseinner automorphism of a right angled Artin group is inner.

Let Γ be a finite graph (without loops or multiple edges) with the set of vertices V .For any vertex v ∈ V its star star(v) consists of all vertices (including v itself) that areadjacent to v in Γ. If S ⊆ V , then star(S) :=

⋂v∈S star(v). Observe that for two subsets

S, T ⊆ V , T ⊆ star(S) happens if and only if S ⊆ star(T ).

18 ASHOT MINASYAN

Let G = G(Γ) be the associated right angled Artin group. To simplify notation, wewill identify elements of V with the corresponding generators of G. Then for each v ∈ V ,star(v) contains precisely those elements from V that commute with v in G. For anysubset A of G, A±1 will denote the union A∪A−1 ⊆ G. Thus every element g ∈ G can berepresented as a word W in letters from V±1. The support supp(W ) is the set of all v ∈ Vsuch that v±1 appears as a letter in W . A word W is said to be graphically reduced if it hasno subwords of the form vUv−1 or v−1Uv, where v ∈ V and supp(U) ⊆ star(v). Evidently,if the word W is not graphically reduced, then one can find a shorter word representingthe same element of the group G. This process will eventually terminate (because thelength of W is finite), hence for each element g ∈ G there exists a graphically reducedword representing it in G.

E. Green [28] proved that if two graphically reduced words W and W ′ represent thesame element g ∈ G, then W and W ′ have the same length and supp(W ) = supp(W ′).Moreover, for any given g ∈ G, graphically reduced words are precisely the shortestpossible words representing g in G (proofs of these facts using re-writing systems can alsobe found in [24, Sec. 2.2]). Therefore, for any element g we can define its length |g| as thelength of any graphically reduced word W representing g in G, and its support supp(g)as supp(W ).

Finally, for any g ∈ G define FL(g) – the set of first letters of g – as the set of allletters a ∈ V±1 such that a appears as the first letter of some graphically reduced wordW representing g in G. Similarly, we define the set of last letters LL(g) of g as thosea ∈ V±1 that appear as a last letter of some graphically reduced word representing g inG. A useful fact observed by Green in [28] states that for any g ∈ G the letters in FL(g)pairwise commute (in G). Evidently, LL(g) = (FL(g−1))−1.

Consider any subset S of V and let ∆ be the full subgraph of Γ on the vertices from S.Let H denote the right angled Artin group corresponding to ∆. The identity map on Scan be regarded as a map from the generating set of H into G. Since ∆ is a subgraph of Γall the relations between these generators of H hold between their images in G. Therefore,by von Dyck’s Theorem, there is a homomorphism ξ : H → G extending the identity mapon S.

On the other hand, since ∆ is a full subgraph of Γ, by von Dyck’s Theorem, the mapρS : V t 1 → V t 1 defined by ρS(1) := 1 and

(6.1) ρS(v) :=

v if v ∈ S1 if v ∈ V \ S ,

can be extended to a homomorphism ρH : G → H. Obviously, the composition ρH ξ :H → H is the identity map on H. Therefore ξ is injective, hence it is an isomorphismbetween H and the subgroup of G generated by S. Consequently, ρH , regarded as anendomorphism of G, becomes a (canonical) retraction of G onto 〈S〉 ≤ G.

For any S ⊆ V the subgroup H := 〈S〉 ≤ G is called special (or full, or canonicallyparabolic, depending on the source). Note that the trivial subgroup 1 ≤ G is also specialand corresponds to the empty subset of V . As we saw above, any special subgroup is a


right angled Artin group itself, and is a retract of G. It is easy to see that if S, T are twosubsets of V then the corresponding maps ρS and ρT defined by (6.1) commute with eachother. This leads to the following important observation.

Remark 6.1. If H and F are special subgroups of a right angled Artin group G, then Hand F are retracts of G and the corresponding canonical retractions ρH , ρF ∈ End(G)commute.

Remark 6.2. If S, T ⊆ V then 〈S〉 ∩ 〈T 〉 = 〈S ∩ T 〉.

Indeed, by Remarks 6.1 and 4.2, we have

〈S〉 ∩ 〈T 〉 = ρ〈T 〉(〈S〉) = 〈ρ〈T 〉(S)〉 = 〈ρT (S)〉 = 〈S ∩ T 〉.

Recall that a group G is said to have the Unique Root property if for any positive integern and arbitrary elements x, y ∈ G the equality xn = yn implies x = y in G. The groupG is called bi-orderable if G can be endowed with a total order , which is bi-invariant,i.e., for any x, y, z ∈ G, if x y, then zx zy and xz yz.

Lemma 6.3. Right angled Artin groups have the Unique Root property.

Proof. G. Duchamp and D. Krob [20] (see also [21]) proved that right angled Artin groupsare bi-orderable. Let G be a right angled Artin group, and let be a total bi-invariantorder on G.

Suppose that xn = yn for some x, y ∈ G, n ∈ N, and x 6= y. Without loss of generalitywe can assume that x ≺ y. Let us show that xk ≺ yk for every k ∈ N. This is true fork = 1, so, proceeding by induction on k, suppose that k ≥ 2 and xk−1 ≺ yk−1 has alreadybeen shown. Then xk = xk−1x ≺ xk−1y ≺ yk−1y = yk, where we used the inductionhypothesis together with the bi-invariance of the order.

Hence, we have proved that xn ≺ yn, contradicting to xn = yn. Thus x = y.

The Unique Root property for right angled Artin groups can also be easily establishedusing the fact that these groups are residually torsion-free nilpotent, which was also provedin [20].

Lemma 6.4. Let H be a conjugate of a special subgroup in a right angled Artin group G.If K ≤ G is a subgroup such that |K : (K ∩H)| <∞ then K ⊆ H.

Proof. By the assumptions, H is a retract of G. Let ρH ∈ End(G) denote the correspond-ing retraction. Take any x ∈ K. Since |K : (K ∩ H)| < ∞, there is n ∈ N such thatxn ∈ H. Therefore, setting y := ρH(x) ∈ H, we obtain xn = ρH(xn) = yn. And theUnique Root property for G implies that x = y ∈ H. Thus K ⊆ H.

After writing down the proof of the next technical result (Lemma 6.5), the authorlearned that it has already been established by A. Duncan, I. Kazachkov and V. Remeslen-nikov in their recent paper [22, Prop. 2.6]. However, the proof presented here is somewhatdifferent, and the author decided to keep it in this work for completeness.

20 ASHOT MINASYAN

Lemma 6.5. Let G be a right angled Artin group associated to a finite graph Γ withvertex set V. Suppose that S, T ⊆ V and g ∈ G. Then there are P ⊆ T and h ∈ 〈T 〉such that g〈S〉g−1 ∩ 〈T 〉 = h〈P〉h−1 in G. Consequently, the intersection of conjugates oftwo special subgroups in G is a conjugate of a special subgroup of G.

Proof. We will use induction on (|S| + |g|), where |S| denotes the cardinality of S. IfS = ∅, then 〈S〉 = 1 and the statement is trivial. If |g| = 0, i.e., g = 1 in G, theng〈S〉g−1 ∩ 〈T 〉 = 〈S〉 ∩ 〈T 〉 = 〈S ∩ T 〉 by Remark 6.2.

Thus we can assume that S 6= ∅ and g 6= 1, n := |S|+ |g| ≥ 2, and the claim has beenproved for all S and g with |S|+ |g| < n.

If there is a ∈ FL(g) ∩ T ±1, set f := a−1g. Then |f | < |g|, a〈T 〉a−1 = 〈T 〉 and

g〈S〉g−1 ∩ 〈T 〉 = a(f〈S〉f−1 ∩ 〈T 〉

)a−1 = (ah)〈P〉(ah)−1

for some h ∈ 〈T 〉 and some P ⊆ T by the induction hypothesis.

If, on the other hand, there is b ∈ LL(g) ∩(S ∪ star(S)

)±1, set f := gb−1. Then

|f | < |g|, b〈S〉b−1 = 〈S〉 and g〈S〉g−1 ∩ 〈T 〉 = f〈S〉f−1 ∩ 〈T 〉 and we can apply theinduction hypothesis once again.

Therefore, we can suppose that FL(g)∩T ±1 = ∅ and LL(g)∩(S ∪ star(S)

)±1= ∅. We

assert that in this case

(6.2) g〈S〉g−1 ∩ 〈T 〉 =⋃s∈S

(g 〈S \ s〉 g−1 ∩ 〈T 〉

).

Indeed, if (6.2) is false, then there exist x ∈ 〈S〉 and y ∈ 〈T 〉 such that supp(x) = S andgxg−1 = y in G.

Choose graphically reduced words W,X and Y representing in G the elements g, x andy respectively, so that supp(X) = S and supp(Y ) ⊆ T . Let a be the first letter of W , thena = v±1 for some v ∈ V . According to our assumptions, v ∈ supp(WXW−1)\supp(Y ) andWXW−1 = Y in G. Hence the left-hand side of the latter equality cannot be graphicallyreduced.

Note that no letter a of W can be cancelled with a letter of X in the word WXW−1,because this would mean that a ∈ supp(X)±1 = S±1 and a commutes with the suffix ofW after it, hence a ∈ LL(g) ∩ S±1 = ∅. Similarly, no letter from X can cancel with aletter from W−1, therefore a reduction in WXW−1 can occur only from the presence ofa subword cUc−1, where c is a letter from the initial copy of W and U contains X as asubword. Thus, c = w±1 for some w ∈ V , supp(W ) ⊆ supp(U) ⊆ star(w). Consequently,S ⊆ star(w), and c is a last letter of g, because it commutes with the suffix of W after it.This implies that c ∈ LL(g) ∩ star(S)±1 6= ∅ contradicting to our assumption.

Therefore (6.2) is true, implying that the group K := g〈S〉g−1 ∩ 〈T 〉 ≤ G is covered bya finite union of its subgroups. A classical theorem of B. Neumann [45] claims that in thiscase one of these subgroups must have finite index in K. Thus there is s0 ∈ S such that|K : (g 〈S \ s0〉 g−1 ∩ 〈T 〉) | < ∞. Using the induction hypothesis, we can find P ⊆ Tand h ∈ 〈T 〉 such that g 〈S \ s0〉 g−1 ∩ 〈T 〉 = h〈P〉h−1. Therefore h〈P〉h−1 ≤ K and


|K : h〈P〉h−1| < ∞. Hence Lemma 6.4 can be applied to achieve the required equalityK = h〈P〉h−1.

Let us recall a few more facts about right angled Artin groups.

An element g ∈ G is said to be A-cyclically reduced if it cannot be written as g = aha−1,where a ∈ V±1 and |h| = |g| − 2 (we have added the letter “A” to avoid confusion witha similar notion for special HNN-extensions introduced in Section 7). In the paper [56]H. Servatius proved that for every element g of a right angled Artin group G there existsa unique A-cyclically reduced element h such that g = fhf−1 for some f ∈ G with|g| = |h|+ 2|f |. In particular, supp(h) ⊆ supp(g). Therefore, if g ∈ G is not A-cyclicallyreduced then |g2| = |fh2f−1| ≤ 2|h|+ 2|f | < 2|g|. Thus we obtain the following

Remark 6.6. If an element g ∈ G is not A-cyclically reduced, then for any graphicallyreduced word W representing g in G, the word W 2 ≡ WW cannot be graphically reduced.

Another consequence of the above theorem of Servatius is that every given element ofG is conjugate to a unique (up to a cyclic permutation) A-cyclically reduced element. Inparticular, we can make

Remark 6.7. If the elements g, h ∈ G are A-cyclically reduced and conjugate in G, thensupp(g) = supp(h).

A special subgroup A of the right angled Artin group G is said to be maximal if A = 〈S〉for some maximal proper subset S of V (i.e., if |S| = |V| − 1).

Lemma 6.8. For any non-trivial element g ∈ G there is a maximal special subgroup Ain G such that g /∈ AG.

Proof. Arguing by contradiction, suppose that there are f1, . . . , fn ∈ G such that g ∈⋂ni=1 fiAif

−1i in G, where A1, . . . , An is the list of all maximal special subgroups of G.

Then for each i ∈ 1, . . . , n, there is an A-cyclically reduced element ai ∈ Ai \ 1 suchthat g is conjugate to ai in G. Choose any letter v ∈ supp(a1) and take j ∈ 1, . . . , nsuch that Aj = 〈V \ v〉. Then a1 must be conjugate to aj in G, which is impossible byRemark 6.7, because v ∈ supp(a1) \ supp(aj) (as supp(aj) ⊆ V \ v). This contradictionproves the lemma.

Recall that an automorphism φ of a group G is called pointwise inner if for each g ∈ Gthere exists f = f(g) ∈ G, such that φ(G) = fgf−1 in G. Let Autpi(G) denote the setof all pointwise inner automorphisms of G. It is easy to see that Autpi(G) is a normalsubgroup of the full automorphism group Aut(G), containing the subgroup of all innerautomorphisms Inn(G).

We are now going to prove that the the group of pointwise inner automorphisms of aright angled Artin group G coincides with the group of inner automorphisms of G.

Proposition 6.9. For any right angled Artin group G we have Autpi(G) = Inn(G).

22 ASHOT MINASYAN

Proof. Suppose there exists φ ∈ Autpi(G) \ Inn(G). Since φ maps every generator of Gto its conjugate, we can replace φ by its composition with an inner automorphism of Gto assume that φ(v) = v for some v ∈ V .

Then there is an automorphism µ lying in the right coset Inn(G)φ ⊂ Autpi(G) suchthat the set Fix(µ) is maximal, where Fix(µ) denotes the subset of all elements in V fixedby µ. Note that Fix(µ) $ V because µ 6= idG (the identity map idG of G does not belongto the coset Inn(G)φ by our assumption). And Fix(µ) 6= ∅ since v ∈ Fix(φ) 6= ∅.

Let Fix(µ) = v1, . . . , vk ⊂ V and pick any w ∈ V \Fix(µ). By the assumptions, thereis f ∈ G such that µ(w) = fwf−1. Choose a shortest element f with this property.

First, suppose that there is a ∈ FL(f) ∩(⋂k

i=1 star(vi))±1

. Then a ∈⋂ki=1CG(vi),

hence after defining a new automorphism λ ∈ Inn(G)φ by λ(g) := a−1µ(g)a for all g ∈ G,we will have Fix(µ) ⊆ Fix(λ), λ(w) = (a−1f)w(a−1f)−1 and |a−1f | < |f |. Note thata−1f 6= 1 because, otherwise, we would have w ∈ Fix(λ) contradicting to the maximalityof Fix(µ). Thus we can replace µ with λ, making f shorter. We can continue doing thesame for λ, and so on. Eventually we will end up with an automorphism µ ∈ Inn(G)φ(we keep the same notation for it, although the actual automorphism may be different)such that µ(w) = fwf−1, f is a shortest element with this property, f 6= 1, and

(6.3) FL(f) ∩ star (Fix(µ))±1 = FL(f) ∩

(k⋂i=1

star(vi)

)±1

= ∅.

Denote S := Fix(µ) ∩ FL(f)±1 ⊂ V . Using (6.3), we see that for every s ∈ S there isv(s) ∈ Fix(µ) such that v(s) /∈ star(s). Note that in this case v(s) /∈ FL(f)±1 becauses ∈ FL(f)±1 and any two elements of FL(f)±1 commute. Set T := v(s) | s ∈ S ⊆Fix(µ) \ FL(f)±1, and write Fix(µ) = t1, . . . , tl t s1, . . . , sm, where T = t1, . . . , tl(consequently, S ⊆ s1, . . . , sm). Finally, define the words T := t1 . . . tl, S := s1 . . . smand let g be the element represented by the word by TSTw in G.

Since µ ∈ Autpi(G), there exists x ∈ G such that µ(g) = xgx−1. On the other hand,µ(g) = TSTUwU−1, for some graphically reduced non-empty word U representing f inG. Note that the word UwU−1 is graphically reduced (otherwise, we could make U , andhence f , shorter) and w /∈ supp(TST ) = Fix(µ). Therefore the only possible reductionwhich could occur in the word W ≡ TSTUwU−1 would arise from cancellation of a letterfrom TST with a letter from U or U−1. However, no letter t from the first copy of Tcould cancel with a letter from U or U−1 in W , because supp(S) 6⊂ star(t) (as t = v(s)for some s ∈ S and s ∈ supp(S) \ star(v(s))). On the other hand, if some letter t fromthe second copy of T in W cancelled with some letter a from U , then a ∈ FL(f), but thiswould contradict to t ∈ T and T ∩ FL(f)±1 = ∅. If this letter t cancelled with a letter afrom U−1 in W , then we would have supp(U) ⊂ star(t), which is impossible as t = v(s)for some s ∈ supp(U) and s /∈ star(v(s)). Therefore, if W is not graphically reduced,then a letter s from S must cancel with a letter a = s−1 from U or U−1, in particular,T = supp(T ) ⊆ star(s), implying that s /∈ S. However, if s cancels with a letter fromU , we see that a ∈ FL(f), hence s ∈ S, which is false. And if s cancelled with a letter


a from U−1, then we would get a contradiction with the fact that UwU−1 is graphicallyreduced, because a−1 = s is a letter of U lying between s and a in W .

Therefore W ≡ TSTUwU−1 is a graphically reduced word representing µ(g) in G. Con-sequently, |µ(g)| = 2‖T‖+‖S‖+2‖U‖+1 > 2‖T‖+‖S‖+1 = |g| since ‖U‖ > 0 (‖U‖ de-notes the length of the word U). But µ(g) = xgx−1, hence µ(g) is not A-cyclically reduced.By Remark 6.6, a reduction can be made in the word W 2 ≡ TSTUwU−1TSTUwU−1.But an argument similar to the above shows that this is impossible.

Thus we have arrived to a contradiction, which proves that Autpi(G) = Inn(G), asneeded.

Remark 6.10. The reader could have noticed that in the proof of Proposition 6.9 we haveactually shown more than it claims. In fact, we have proved that any endomorphism φof a right angled Artin group G, which maps each conjugacy class of G into itself, is aninner automorphism of G.

7. Special HNN-extensions

The purpose of this section is to develop necessary tools for dealing with special HNN-extensions.

Let A be a group and let H ≤ A be a subgroup.

Definition 7.1. The special HNN-extension of A with respect to H is the group G givenby the presentation

(7.1) G = 〈A, t ‖ tht−1 = h for every h ∈ H〉.

In other words, the special HNN-extension G is a particularly simple HNN-extension ofA, where both of the associated subgroups are equal to H and the isomorphism betweenthese subgroups is the identity map on H.

Let Γ be a finite graph with the set of vertices V of cardinality n ∈ N . The reason whywe are interested in special HNN-extensions is the the observation below.

Remark 7.2. Let G be the right angled Artin group associated to Γ. Then G can beobtained from the trivial group via a sequence of special HNN-extensions. More precisely,there are right angled Artin groups 1 = G0, G1, . . . , Gn = G such that Gi+1 is aspecial HNN-extension of Gi with respect to some special subgroup Hi ≤ Gi for everyi = 0, . . . , n− 1.

The groups Gi can be constructed as follows. Let V = v1, . . . , vn and denote Si+1 :=v1, . . . , vi ∩ star(vi+1) ⊂ V for i = 1, . . . , n − 1. Set G0 := 1, G1 := 〈v1〉 (the infinitecyclic group generated by v1), and

Gi+1 := 〈Gi, vi+1 ‖ vi+1sv−1i+1 = s for every s ∈ Si+1〉, i = 1, . . . , n− 1.

Clearly, G = Gn and for each i, Gi+1 is a special HNN-extension of Gi with respect to thespecial subgroup 〈Si+1〉 of Gi, and Gi is a special subgroup of G generated by v1, . . . , vi.

24 ASHOT MINASYAN

Remark 7.3. If G is a right angled Artin group associated to Γ, then for any maximalspecial subgroup A ≤ G, G splits as a special HNN -extension (7.1) of A with respect toa certain special subgroup H of A.

Indeed, if A = 〈S〉, where S ⊂ V and V = S t t, set U := star(t) \ t ⊆ S. ThenG = 〈A, t ‖ tut−1 = u, ∀u ∈ U〉 is a special HNN-extension of A with respect to thesubgroup H := 〈U〉 ≤ A.

Special HNN-extensions are usually much easier to deal with than general HNN-extensions. Throughout this section we will limit ourselves to considering only the formerones, even though most of the statements can be re-formulated in the general situation.

Let G be the special HNN-extension given by (7.1). von Dyck’s Theorem yields thefollowing Universal Property of special HNN-extensions, which will be important for us:

Remark 7.4. For any group B, every homomorphism ψ : A→ B can be naturally extendedto a homomorphism ψ : G→ P , where P := 〈B, s ‖ sxs−1 = x, ∀x ∈ ψ(H)〉 is the special

HNN-extension of B with respect to ψ(H), so that ψ|A = ψ and ψ(t) = s.

Lemma 7.5. In the notations of Remark 7.4, ker(ψ) = N , where N C G is the normalclosure of ker(ψ) ≤ A ≤ G in G.

Proof. Obviously, N ≤ ker(ψ), and hence N ∩ A = ker(ψ) ∩ A = ker(ψ). Let φ : G →Q := G/N be the natural epimorphism with ker(φ) = N . Consequently, if we define

θ : Q → P to be the natural epimorphism with the kernel φ(ker(ψ)), then we will have

ψ = θ φ.

Observe that φ(t)φ(x)φ(t)−1 = φ(txt−1) = φ(x) in Q for every x ∈ H, and the mapξ : ψ(A) → φ(A) defined by ξ(ψ(a)) := φ(a) for all a ∈ A, is an isomorphism, sinceker(ψ) = ker(φ) ∩ A. Therefore, by von Dyck’s Theorem, there is a homomorphism

ξ : P → Q such that ξ(ψ(a)) = φ(a) for every a ∈ A and ξ(s) = φ(t). It is easy to see

that ξ θ : Q→ Q is the identity map on Q. Hence, θ is injective, that is 1 = ker(θ) =

φ(ker(ψ)) in Q, implying that ker(ψ) ≤ ker(φ) = N . Thus ker(ψ) = N .

Lemma 7.6. Suppose that ρ ∈ End(A) is a retraction of A onto its subgroup B. Thenthere are retractions ρ1, ρ2 ∈ End(G) of G onto subgroups B ≤ G and C := 〈B, t〉 ≤ Grespectively, such that ρi|A = ρ for i = 1, 2, ρ1(t) = 1 and ρ2(t) = t.

Proof. Define the maps ρi : A t t → A t t by ρi(a) := ρ(a) for each a ∈ A, i = 1, 2,and ρ1(t) := 1 and ρ2(t) := t.

The map ρ1 can be extended to an endomorphism ρ1 : G→ G by von Dyck’s Theorem,and the map ρ2 can be extended to an endomorphism ρ2 : G → G by Remark 7.4.Obviously, ρ1 and ρ2 are retractions of G onto B and C respectively.

Every element of the special HNN-extension G, given by (7.1), is a product of the form

(7.2) x0tε1x1t

ε2 . . . tεnxn

for some n ∈ N ∪ 0, xi ∈ A, i = 0, . . . , n, and εj ∈ Z \ 0, j = 1, . . . , n.


The product (7.2) is said to be reduced, if xi /∈ H for every i ∈ 1, 2, . . . , n − 1.Since t commutes with every element of H, it is easy to see that any g ∈ G is equal tosome reduced product in G. By Britton’s Lemma (see [37, IV.2]) a non-empty reducedproduct represents a non-trivial element in G. It follows, that if two reduced productsx0t

ε1x1tε2 . . . tεnxn and y0t

ζ1y1tζ2 . . . tζmym are equal in G, then m = n and εi = ζi for

every i = 1, . . . , n (see [37, IV.2.3]).

Suppose that an element g ∈ G is equal to a product tε1x1tε2 . . . tεnxn. Let us fix this

presentation for g. Any product of the form tεkxktεk+1 . . . tεnxnt

ε1x1 . . . tεk−1xk−1 ∈ G,

for some k ∈ 1, . . . , n, is said to be a cyclic permutation of g. The element g =tε1x1t

ε2 . . . tεnxn is called cyclically reduced if each of its cyclic permutations is reduced. Aprefix of g is an element of the form tε1x1t

ε2 . . . tεkxk for some k ∈ 0, 1, . . . , n (if k = 0then we have the empty prefix, corresponding to the trivial element of G). Similarly, asuffix of g is an element of the form tεlxl . . . t

εnxn for some l ∈ 1, 2, . . . , n+ 1.It is not difficult to see that every element f ∈ G either belongs to AG in G or is

conjugate to some non-trivial cyclically reduced element in G.

Below is the statement of Collins’s Lemma (see [37, IV.2.5]) in the case of specialHNN-extensions.

Lemma 7.7. Suppose that g = tε1x1tε2 . . . tεnxn and f = tζ1y1t

ζ2 . . . tζmym are cyclicallyreduced in G, with n ≥ 1. Then g /∈ AG. And if f is conjugate to g in G then m = n andthere exist h ∈ H and a cyclic permutation f ′ of f such that f ′ = hgh−1 in G.

We will also use the following description of centralizers in special HNN-extensions:

Proposition 7.8. Let G be the special HNN-extension given by (7.1). Suppose that theelement g = tε1x1t

ε2 . . . tεnxn ∈ G is cyclically reduced and n ≥ 1.

If xn ∈ H then n = 1 and CG(g) = 〈t〉 × CH(x1) = 〈t〉 × CH(g) ≤ G.

If xn ∈ A \H, let p1, . . . , pk, 1 ≤ k ≤ n + 1, be the set all of prefixes of g satisfyingp−1i gpi ∈ gH in G. For each i = 1, . . . , k, choose hi ∈ H such that hip

−1i gpih

−1i = g, and

define the finite subset Ω ⊂ G by Ω := hip−1i | i = 1, . . . , k. Then CG(g) = CH(g)〈g〉Ω.

In order to prove Proposition 7.8 we will need two lemmas below. The proof of thenext statement is similar to the proof of Collins’s Lemma.

Lemma 7.9. Let g = tε1x1tε2 . . . tεnxn and f = tζ1y1t

ζ2 . . . tζnyn be cyclically reducedelements of G, with n ≥ 1 and xn /∈ H. Assume that cgc−1 = f in G, where the elementc is equal to the reduced product z0t

η1z1tη2 . . . tηmzm. Then there are the following three

mutually exclusive possibilities.

a) m = 0 and c ∈ H;b) m ≥ 1, zm ∈ H and there is a prefix p of g such that c = hp−1gl in G, for some

h ∈ H and l ∈ Z, l ≤ 0;c) m ≥ 1, xnz

−1m ∈ H and there is a suffix s of g such that c = hsgl in G, for some

h ∈ H and l ∈ Z, l ≥ 0.

26 ASHOT MINASYAN

Proof. First, suppose that m = 0, i.e., c = z0 ∈ A. Then f−1cgc−1 = 1 in G. Yielding

y−1n t−ζn . . . y−1

1 t−ζ1z0tε1x1t

ε2 . . . tεnxnz−10 = 1.

Therefore the left-hand side is not reduced, hence c = z0 ∈ H.

Assume now that m ≥ 1. The equality cgc−1 = f in G gives rise to the equation

z0tη1 . . . tηmzmt

ε1x1tε2 . . . tεnxnz

−1m t−ηm . . . t−η1z−1

0 = tζ1y1 . . . tζnyn.

The left-hand side cannot be reduced because it contains more t-letters than the right-hand side. Hence either zm ∈ H or xnz

−1m ∈ H (note that both of these inclusions cannot

happen simultaneously since xn /∈ H). Let us consider the case when zm ∈ H, as thesecond case is similar. Then tηmzmt

ε1 = zmtηm+ε1 and, thus, we get

(7.3) z0tη1 . . . tηm−1(zm−1zm)tηm+ε1x1t

ε2 . . . tεnxnz−1m t−ηmz−1

m−1 . . . t−η1z−1

0 = tζ1y1 . . . tζnyn.

Once again, we see that the left-hand side of the above equation cannot be reduced. Inthe case when m = 1, this implies that ηm + ε1 = 0. On the other hand, if m > 1, thenzm−1 ∈ A \ H, hence zm−1zm /∈ H, and again, in order for a reduction to be possiblewe must have ηm + ε1 = 0. Hence ε1 = −ηm and z−1

m t−ηmz−1m−1 = tε1x1h

−11 , where

h1 := zm−1zmx1 ∈ A. And (7.3) becomes

(7.4) (z0tη1 . . . zm−2t

ηm−1h1)(tε2x2 . . . tεnxnt

ε1x1)(z0tη1 . . . tηm−1h1)−1 = tζ1y1 . . . t

ζnyn.

Now, if m = 1, i.e., c = z0t−ε1z1 = z0z1t

−ε1 , then we have

1 = f−1cgc−1 = (y−1n t−ζn . . . y−1

1 t−ζ1)h1(tε2x2 . . . tεnxnt

ε1x1)h−11 .

Hence h1 ∈ H and c = h1(tε1x1)−1, where tε1x1 is a prefix of g.

Otherwise, if m = M ≥ 2 we will use induction on m to prove the claim b) of the lemma.Thus, we will assume that b) has been established for all elements c with 1 ≤ m ≤M−1.

Note that x1h−11 = z−1

m z−1m−1 /∈ H since zm ∈ H and zm−1 ∈ A \H (because m ≥ 2 and

the product z0tη1 . . . tηm−1zm−1t

ηmzm was assumed to be reduced).

Let us look at the equation (7.4). Since m−1 ≥ 1, the left-hand side cannot be reduced.And a reduction in it can only occur if h1 ∈ H because x1h

−11 /∈ H. Hence we are in

the case b) of the lemma, and can apply the induction hypothesis to (7.4). Thus thereis a prefix p of the element g1 := tε2x2 . . . t

εnxntε1x1, h ∈ H and l ∈ Z, l ≤ 0, such that

z0tη1 . . . zm−2t

ηm−1h1 = hp−1gl1. Consequently,

c = z0tη1 . . . tηm−1zm−1t

ηmzm = z0tη1 . . . tηm−1h1x

−11 t−ε1 = hp−1gl1x

−11 t−ε1 = hq−1gl,

where q = tε1x1p. It is easy to see that either q is a prefix of g, or q = gtε1x1. In thelatter case, c = h(tε1x1)−1gl−1 and tε1x1 is a prefix of g. Thus the step of induction isestablished, and the proof of the lemma is finished.

The next lemma treats the case which was not covered by Lemma 7.9.

Lemma 7.10. Suppose that g = tεx ∈ G, where ε ∈ Z \ 0 and x ∈ H. Then CG(g) =〈t〉 × CH(x) = 〈t〉 × CH(g). In particular, CG(tε) = 〈t〉 ×H ≤ G.


Proof. For any c ∈ CG(g) we have cgc−1 = g. Let z0tη1z1t

η2 . . . tηmzm be a reduced productrepresenting c in G. Then we have

(7.5) z0tη1z1t

η2 . . . tηmzmtεxz−1

m t−ηm . . . t−η2z−11 t−η1z−1

0 = tεx.

Arguing as in the proof of Lemma 7.9, we see that if m = 0 then c = z0 ∈ H andtεx = z0t

εxz−10 = tεz0xz

−10 , thus 1 = z0xz

−10 , i.e., c = z0 ∈ CH(x).

So, assume now that m ≥ 1. Then the equation (7.5) implies that either zm ∈ Hor xz−1

m ∈ H. But either of these inclusions leads to zm ∈ H because x ∈ H by theassumptions. Therefore tηmzmt

εxz−1m t−ηm = zmxz

−1m tε in G and (7.5) becomes

z0tη1 . . . zm−2t

ηm−1(zm−1zmxz−1m )tεz−1

m−1tηm−1z−1

m−2 . . . t−η1z−1

0 = tεx.

If m ≥ 2, then zm−1 ∈ A\H, hence zm−1zmxz−1m /∈ H, and the above equation contradicts

to Britton’s Lemma: the left-hand side is reduced, but contains more t-letters than theright-hand side.

Therefore m = 1, i.e., c = z0tη1z1 and z1 ∈ H. Consequently,

1 = g−1cgc−1 = x−1t−εz0tη1z1t

εxz−11 t−η1z−1

0 = x−1t−ε(z0z1xz−11 )tεz−1

0 .

Applying Britton’s Lemma again, we achieve z0z1xz−11 ∈ H, implying that z0 ∈ H and

1 = x−1z0z1xz−11 z−1

0 in G. That is, z0z1 ∈ CH(x), and c = z0tη1z1 = tη1z0z1 ∈ 〈t〉CH(x).

Thus we proved that CG(g) ⊆ 〈t〉CH(x). Finally, since t commutes with every elementfrom H, it is clear that t ∈ CG(g) and CH(x) ⊂ CH(g) ⊂ CG(g). Hence CG(g) =〈t〉CH(x) = 〈t〉 × CH(x) ≤ G. The equality CH(g) = CH(x) is immediate.

Proof of Proposition 7.8. If xn ∈ H, g can be cyclically reduced only when n = 1, andthe claim follows from Lemma 7.10.

So, we can assume that xn ∈ A \ H. Therefore we are able to apply Lemma 7.9 to gand f := g, which tells us that for any c ∈ CG(g) there exist h ∈ H and l ∈ Z such thateither there is a prefix p of g with c = hp−1gl, or there is a suffix s of g with c = hsgl.Note that in the latter case there is a prefix p of g such that s = p−1g. Hence we canassume that c = hp−1gl for some prefix p of g, h ∈ H and l ∈ Z.

But then g = cgc−1 = hp−1gph−1, hence p = pi for some i ∈ 1, . . . , k. The equalitieshp−1

i gpih−1 = g = hip

−1i gpih

−1i yield h−1gh = p−1

i gpi = h−1i ghi. Consequently hh−1

i ∈CH(g) and h ∈ CH(g)hi. Thus c = hp−1

i gl ∈ CH(g)hip−1i gl ⊆ CH(g)Ω〈g〉.

We have shown that CG(g) ⊆ CH(g)Ω〈g〉. Observe that Ω ⊂ CG(g) by definition, henceCG(g) = CH(g)Ω〈g〉 = CH(g)〈g〉Ω.

Now we formulate a criterion for conjugacy in special HNN-extensions.

Lemma 7.11. Let G be the special HNN-extension (7.1). Suppose that B ≤ A is asubgroup and g, f ∈ G are elements represented by reduced products x0t

ε1x1tε2 . . . tεnxn

and y0tζ1y1t

ζ2 . . . tζmym respectively, with n ≥ 1. Then f ∈ gB in G if and only if all ofthe following conditions hold:

(i) m = n and εi = ζi for all i = 1, . . . , n;

28 ASHOT MINASYAN

(ii) y0y1 . . . yn ∈ (x0x1 . . . xn)B in A;(iii) for every b0 ∈ B with y0y1 . . . yn = b0(x0x1 . . . xn)b−1

0 in A, the intersection I ⊆ Ais non-empty, where

I := b0CB(x0 . . . xn) ∩ y0Hx−10 ∩ (y0y1)H(x0x1)−1 ∩ · · · ∩ (y0 . . . yn−1)H(x0 . . . xn−1)−1.

Proof. First we establish the sufficiency. Assume that the conditions (i),(ii) and (iii)hold. Take any b0 ∈ B satisfying (iii) (it exists by (ii)) and let I be the correspondingintersection. Then there exists an element b ∈ I. The inclusion b ∈ b0CB(x0 . . . xn) impliesthat y0y1 . . . yn = b(x0x1 . . . xn)b−1. We will show that f−1bgb−1 = 1, which is equivalent(in view of (i)) to

(7.6) y−1n t−εn . . . t−ε2y−1

1 t−ε1y−10 bx0t

ε1x1tε2 . . . tεnxnb

−1 = 1.

Note that since b ∈ I, y−10 bx0 ∈ H, hence y−1

1 t−ε1y−10 bx0t

ε1x1 = y−11 y−1

0 bx0x1 ∈H. Therefore we can continue reducing the left-hand side of (7.6), until it becomesy−1n . . . y−1

1 y−10 bx0x1 . . . xnb

−1, which is equal to 1 in G. Thus the sufficiency is proved.

To obtain the necessity, assume that f = bgb−1 for some b ∈ B, that is,

bx0tε1x1t

ε2 . . . tεnxnb−1 = y0t

ζ1y1tζ2 . . . tζmym.

Since both of the sides of the above equality are reduced, applying Britton’s Lemma weobtain (i). Therefore the equation (7.6) holds in G.

Note that there is a canonical retraction ρA ∈ End(G) of G onto A, such that ρA(t) = 1(apply Lemma 7.6 to the identity map on A). Hence, y0 . . . yn = ρA(f) = ρA(bgb−1) =b(x0 . . . xn)b−1, yielding (ii).

To achieve (iii), take any b0 ∈ B satisfying y0y1 . . . yn = b0(x0x1 . . . xn)b−10 . Then

b−10 b ∈ CB(x0 . . . xn), i.e., b ∈ b0CB(x0 . . . xn). By Britton’s Lemma the left-hand side of

(7.6) cannot be reduced, therefore y−10 bx0 ∈ H and b ∈ y0Hx

−10 . Consequently, if n ≥ 2,

y−11 t−ε1y−1

0 bx0tε1x1 = y−1

1 y−10 bx0x1 and (7.6) becomes

y−1n t−εn . . . t−ε2(y−1

1 y−10 bx0x1)tε2 . . . tεnxnb

−1 = 1.

Applying Britton’s Lemma to the above equation, we see again that y−11 y−1

0 bx0x1 ∈ H,hence b ∈ (y0y1)H(x0x1)−1. Clearly, we can continue this process, showing that b ∈(y0 . . . yi)H(x0 . . . xi)

−1 for every i ∈ 0, . . . , n − 1. Thus, b ∈ I 6= ∅ and the condition(iii) is satisfied.

Next comes a similar statement about centralizers.

Lemma 7.12. Let G be the special HNN-extension (7.1). Suppose that B ≤ A is asubgroup, and an element g ∈ G is represented by a reduced product x0t

ε1x1tε2 . . . tεnxn in

G, with n ≥ 1. Then the equality CB(g) = I holds in G, where

I := CB(x0 . . . xn) ∩ x0Hx−10 ∩ (x0x1)H(x0x1)−1 ∩ · · · ∩ (x0 . . . xn−1)H(x0 . . . xn−1)−1.

Proof. Basically, we have already shown this while proving Lemma 7.11. Indeed, denotef := g. For any b ∈ I, if we take b0 = 1, the proof of sufficiency in Lemma 7.11 asserts


that bgb−1 = g, i.e., b ∈ CB(g). On the other hand, if bgb−1 = g for some b ∈ B, then theproof of necessity in Lemma 7.11 claims that b ∈ I. Therefore CB(g) = I.

8. Proof of the main result

Throughout this section we assume that G is a right angled Artin group associatedto some fixed finite graph Γ with the set of vertices V . The rank rank(G) of G is, bydefinition, the number of elements in V .

Our proof of the main result will make use of the following two statements below. Thenext Lemma 8.1 was proved by Green in [28]. It can be easily demonstrated by inductionon the number of vertices in the graph associated to a right angled Artin group usingRemark 7.2 and Britton’s Lemma. On the other hand, it also follows from the linearityof such groups, which was established by S. Humphries in [35].

Lemma 8.1. Right angled Artin groups are residually finite.

We are also going to use the following important fact proved by Dyer in [23].

Lemma 8.2. Virtually free groups are conjugacy separable.

The main Theorem 1.1 will be proved by induction on the rank of the right Artin groupG. The lemma below will be used to establish the inductive step:

Lemma 8.3. Assume that every special subgroup B of G satisfies the condition CCG fromDefinition 3.3, and for each g ∈ G, the B-conjugacy class gB is separable in G.

Suppose that A1, . . . , An are special subgroups of G, A0 is a conjugate of a specialsubgroup of G, and b, x0, . . . , xn, y1, . . . , yn ∈ G are arbitrary elements. Then for anyfinite index normal subgroup K C G there exists a finite index normal subgroup L C Gsuch that L ≤ K and

(8.1) bCA0(x0) ∩

n⋂i=1

xiAiyi ⊆ ψ

([bCA0(x0) ∩

n⋂i=1

xiAiyi

]K

)in G/L,

where ψ : G → G/L is the natural epimorphism, b := ψ(b), Ai := ψ(Ai), xi := ψ(xi),i = 0, . . . , n, and yj := ψ(yj), j = 1, . . . , n.

Proof. By the assumptions, A0 = hAh−1 for some special subgroup A of G and someh ∈ G. The proof will proceed by induction on n.

If n = 0, then the existence of a finite index normal subgroup LCG, L ≤ K, enjoying(8.1), follows from the fact that the pair (A0, x0) satisfies the Centralizer Condition CCG,because the pair (A, g) has CCG, where g := h−1x0h ∈ G, and bCA0(x0) = bhCA(g)h−1 inG.

Base of induction: assume n = 1.

Case 1: suppose that bCA0(x0) ∩ x1A1y1 = ∅, which is equivalent to the conditionx1 /∈ bCA0(x0)y−1

1 A1 = bh [CA(g)D]h−1y−11 , where D := (y1h)−1A1(y1h).

30 ASHOT MINASYAN

In this case, by Lemma 6.5, the intersections A∩A1 and A∩D are conjugates of specialsubgroups in G, hence for any f ∈ G, the conjugacy classes fA∩A1 and fA∩D are separablein G (A ∩D = cSc−1 for some special subgroup S of G, hence fA∩D = c

[(c−1fc)S

]c−1),

and the pair (A ∩A1, f) satisfies CCG. Therefore, Lemma 5.7 allows us to conclude thatthe double coset CA(g)D is separable in G. Thus the double coset bCA0(x0)y−1

1 A1 isseparable as well, implying that there is a finite index normal subgroup N CG such thatx1 /∈ bCA0(x0)y−1

1 A1N . After replacing N with N ∩K we can assume that N ≤ K.

Now, since the pair (A0, x0) has CCG, there exists a finite index normal subgroup LCGsuch that L ≤ N ≤ K and ψ−1 (CA0

(x0)) ⊆ CA0(x0)N in G (using the same notations asin the formulation of the lemma). Therefore, ψ−1

(bCA0

(x0)y−11 A1

)⊆ bCA0(x0)y−1

1 A1N ,

yielding that x1 /∈ ψ−1(bCA0

(x0)y−11 A1

). Hence bCA0

(x0) ∩ x1A1y1 = ∅ in G/L, implyingthat (8.1) holds in this first case.

Case 2: bCA0(x0) ∩ x1A1y1 6= ∅ in G.

Let us make the following general observation:

Remark 8.4. Let H,F be subgroups of a group G such that bH ∩ xFy 6= ∅ in G for someelements b, x, y ∈ G. Then for any a ∈ bH ∩ xFy we have bH ∩ xFy = a(H ∩ y−1Fy).

Thus we can pick any a ∈ bCA0(x0)∩x1A1y1, and according to Remark 8.4, we will havebCA0(x0)∩x1A1y1 = a

(CA0(x0) ∩ y−1

1 A1y1

)= aCE(x0) in G, where E := A0 ∩ y−1

1 A1y1 ≤G. By Lemma 6.5, E = cSc−1 for some special subgroup S of G and some c ∈ A0.

As we saw in the beginning of the proof, it follows from our assumptions that the pair(E, x0) satisfies CCG. Hence there must exist a finite index normal subgroup M C Gsuch that M ≤ K and Cϕ(E)(ϕ(x0)) ⊆ ϕ(CE(x0)K) in G, where ϕ : G → G/M denotesthe natural epimorphism. On the other hand, by Lemma 5.6, there is L C G such that|G : L| < ∞, L ≤ M ≤ K and ψ(A) ∩ ψ

((y1h)−1A1y1h1

)⊆ ψ

(A ∩ (y1h)−1A1y1h1

)ψ(M)

in G/L. Therefore

(8.2) A0 ∩ y−11 A1y1 = ψ(h)

[ψ(A) ∩ ψ

((y1h)−1A1y1h1

)]ψ(h−1) ⊆

ψ(h)[ψ(A ∩ (y1h)−1A1y1h1

)ψ(M)

]ψ(h−1) = ψ(A0 ∩ y−1

1 A1y1)ψ(M) = ψ(E)ψ(M)

(in the notations from the formulation of the lemma). Observe that a := ψ(a) ∈bCA0

(x0) ∩ x1A1y1, hence, by Remark 8.4, bCA0(x0) ∩ x1A1y1 = a

(CA0

(x0) ∩ y−11 A1y1

),

and applying (8.2) we achieve

(8.3) bCA0(x0) ∩ x1A1y1 ⊆ aCψ(EM)(x0).

Since L ≤ M , there is an epimorphism ξ : G/L→ G/M such that ker(ξ) = ψ(M) andϕ = ξ ψ. Note that ξ(ψ(EM)) = ξ(ψ(E)ψ(M)) = ϕ(E) and ξ(x0) = ϕ(x0). Considerany z ∈ Cψ(EM)(x0) in G/L. Then

ξ(z) ∈ Cϕ(E)(ϕ(x0)) ⊆ ϕ(CE(x0)K) = ξ(ψ(CE(x0)K)

).

Hence z ∈ ψ(CE(x0)K) ker(ξ) = ψ(CE(x0)K) because ker(ξ) = ψ(M) ≤ ψ(K). Thus wehave shown that Cψ(EM)(x0) ⊆ ψ(CE(x0)K) in G/L. Finally, combining this with (8.3),


we obtain

bCA0(x0) ∩ x1A1y1 ⊆ aψ(CE(x0)K) = ψ(aCE(x0)K) = ψ([bCA0(x0) ∩ x1A1y1]K),

therefore (8.1) holds in Case 2.

Thus we have established the base of induction.

Step of induction: suppose that n ≥ 2 and the statement of the lemma has been

proved for n− 1.

If bCA0(x0) ∩⋂n−1i=1 xiAiyi = ∅ in G, then, by the induction hypothesis, there is a finite

index normal subgroup LCG such that L ≤ K and

bCA0(x0) ∩

n−1⋂i=1

xiAiyi ⊆ ψ

([bCA0(x0) ∩

n−1⋂i=1

xiAiyi

]K

)= ∅ in G/L.

Hence, the left-hand side of (8.1) will also be empty, and thus (8.1) will be true.

Therefore, we can assume that bCA0(x0)∩⋂n−1i=1 xiAiyi 6= ∅ in G. But in this case we can

apply Remark 8.4 (n − 1) times to find some a ∈ G such that bCA0(x0) ∩⋂n−1i=1 xiAiyi =

a(CA0(x0) ∩

⋂n−1i=1 y

−1i Aiyi

)= aCE(x0), where E := A0∩

⋂n−1i=1 y

−1i Aiyi is a conjugate of a

special subgroup in G by Lemma 6.5. Now we can use the base of induction n = 1, to finda finite index normal subgroup MCG such that M ≤ K and for the natural epimorphismϕ : G→ G/M we have

(8.4) ϕ−1(ϕ(a)Cϕ(E)(ϕ(x0)) ∩ ϕ(xnAnyn)

)⊆ [aCE(x0) ∩ xnAnyn]K in G.

By the induction hypothesis, there exists a finite index normal subgroup L C G suchthat L ≤M ≤ K and

(8.5) ψ−1

(bCA0

(x0) ∩n−1⋂i=1

xiAiyi

)⊆

[bCA0(x0) ∩

n−1⋂i=1

xiAiyi

]M in G.

Combining (8.5) with (8.4) and recalling that ker(ψ) = L ≤M = ker(ϕ) we obtain thefollowing in G:

ψ−1

(bCA0

(x0) ∩n⋂i=1

xiAiyi

)= ψ−1

(bCA0

(x0) ∩n−1⋂i=1

xiAiyi

)∩ ψ−1

(xnAnyn

)⊆[

bCA0(x0) ∩n−1⋂i=1

xiAiyi

]M ∩ xnAnynL ⊆ aCE(x0)M ∩ xnAnynM ⊆

ϕ−1(ϕ(a)Cϕ(E)(ϕ(x0))

)∩ ϕ−1 (ϕ(xnAnyn)) = ϕ−1

(ϕ(a)Cϕ(E)(ϕ(x0)) ∩ ϕ(xnAnyn)

)⊆

[aCE(x0) ∩ xnAnyn]K =

[bCA0(x0) ∩

n⋂i=1

xiAiyi

]K.

Hence (8.1) holds in G/L and we have verified the inductive step, finishing the proofof the lemma.

32 ASHOT MINASYAN

The next two statements basically establish the main result. These Lemmas 8.5 and8.6 will be proved by simultaneous induction on rank(G). The proofs of each of these twolemmas when rank(G) = r will use both of their conclusions about right angled Artingroups of ranks strictly less than r.

Lemma 8.5. If G is a right angled Artin group of rank r, then for any special subgroupB of G and any element g ∈ G, the B-conjugacy class gB is separable in G.

Lemma 8.6. Let G be a right angled Artin group of rank r. Then every special subgroupB of G satisfies the Centralizer Condition CCG from Definition 3.3.

The base of induction for both lemmas is r = 0, that is, when G is the trivial group.In this case the two statements are trivial. Therefore the proofs of Lemmas 8.5 and 8.6start with assuming that both of their claims have been established for all right angledArtin groups of rank < r, and will aim to prove the inductive step by considering the casewhen rank(G) = r ≥ 1.

The proofs of Lemmas 8.5 and 8.6 make use of the four auxiliary statements below.These statements – Lemmas 8.7 through 8.10 – start with a right angled Artin group Gof rank r (presented as a special HNN-extension

(8.6) G = 〈A, t ‖ tht−1 = h, ∀h ∈ H〉.of a maximal special subgroup A ≤ G with respect to some special subgroup H ≤ A –see Remark 7.3), and assume that Lemmas 8.5 and 8.6 have already been established forA, since rank(A) = r − 1 < r = rank(G).

Lemma 8.7. Suppose that B is a special subgroup of G contained in A, g ∈ G \ A andf ∈ G \ gB. Then there exists an epimorphism ψ from A onto a finite group Q such that

for the corresponding extension ψ : G→ P from G onto the special HNN-extension P of

Q (with respect to ψ(H)), obtained according to Remark 7.4, we have ψ(f) /∈ ψ(g)ψ(B) inP .

Proof. Let x0tε1x1t

ε2 . . . tεnxn and y0tζ1y1t

ζ2 . . . tζmym be reduced products representing gand f in G respectively. Since g /∈ A we have n ≥ 1.

Case 1: suppose, at first, that the condition (i) from Lemma 7.11 does not hold. ByCorollary 5.4 and Lemma 8.1, the special subgroup H = HH of A is closed in PT (A),hence there is a finite index normal subgroup LCA and the corresponding epimorphismψ : A→ Q := A/L such that such that ψ(xi) /∈ ψ(H), i = 1, . . . , n−1, and ψ(yj) /∈ ψ(H),j = 1, . . . ,m− 1. Let

(8.7) P := 〈Q, s ‖ sqs−1 = q, ∀ q ∈ ψ(H)〉be the special HNN-extension of Q with respect to ψ(H). By Remark 7.4, ψ can be

extended to a homomorphism ψ : G → P such that ψ|A = ψ and ψ(t) = s. Therefore,

ψ(g) = x0sε1x1s

ε2 . . . sεnxn, ψ(f) = y0sζ1 y1s

ζ2 . . . sζm ym and these products are reducedin P , where xi := ψ(xi) and yj := ψ(yj) for all i = 0, . . . , n, j = 0, . . . ,m. And since

the condition (i) did not hold for g and f , this condition will not hold for ψ(g) and ψ(f).

Therefore, ψ(f) /∈ ψ(g)ψ(B) by Lemma 7.11.


Thus we can now assume that n = m and εi = ζi for i = 1, . . . , n. Denote x :=x0 . . . xn ∈ A and y := y0 . . . yn ∈ A.

Case 2: suppose that y /∈ xB in A. Then, by the induction hypothesis, xB is separablein A, hence there is a finite group Q and an epimorphism ψ : A → Q such that ψ(y) /∈ψ(x)ψ(B). Let P be the special HNN-extension of Q defined by (8.7), and let ψ : G→ P

be the corresponding extension of ψ with ψ(t) = s. By Lemma 7.6, there is a retractionρQ ∈ End(P ) of P onto Q (extending the identity map on Q) satisfying ρQ(s) = 1.Therefore, using the above notations, we have

ρQ

(ψ(f)

)= ρQ(y0s

ε1 y1sε2 . . . sεn yn) = y0y1 . . . yn = ψ(y) ∈ Q,

similarly, ρQ

(ψ(g)

)= ψ(x) ∈ Q. And since ρQ

(ψ(B)

)= ψ(B) and ψ(y) /∈ ψ(x)ψ(B) we

can conclude that ψ(f) /∈ ψ(x)ψ(B).

Case 3: we can now assume that both of the conditions (i) and (ii) of Lemma 7.11are satisfied. Choose any b0 ∈ B such that y = b0xb

−10 . As f /∈ gB in G, according to

Lemma 7.11 we must have I = ∅ in A, where

I := b0CB(x) ∩ y0Hx−10 ∩ (y0y1)H(x0x1)−1 ∩ · · · ∩ (y0 . . . yn−1)H(x0 . . . xn−1)−1.

As we saw earlier, H is separable in A, therefore there is a finite index normal subgroupKCA such that xi /∈ HK and yi /∈ HK for 1 ≤ i ≤ n−1. Now, since rank(A) = r−1 < r,the right angled Artin group A satisfies the claims of Lemmas 8.5 and 8.6 by the inductionhypothesis. Consequently, we can apply Lemma 8.3 to A and K, finding a finite indexnormal subgroup LC A such that L ≤ K and

(8.8) b0CB(x) ∩n⋂i=1

xiHyi ⊆ ψ(IK) = ∅ in Q := A/L,

where b0, B, x, xi, H, yi denote the ψ-images of b0, B, x, xi, H, yi in Q respectively. Asbefore we can extend ψ to a homomorphism ψ : G → P , where P is given by (8.7), and

ψ(t) = s. Since L ≤ K we have xi, yi /∈ ψ(H) for i = 1, . . . , n, and so x0sε1x1s

ε2 . . . sεnxnand y0s

ε1 y1sε2 . . . sεn yn are reduced products in P representing the elements ψ(g) and

ψ(f) respectively. Thus, Lemma 7.11, in view of (8.8), implies that ψ(f) /∈ ψ(g)ψ(B) inP . And Lemma 8.7 is proved.

Lemma 8.8. Suppose that g0, f0, f1, . . . , fm ∈ G, and the elements g0 = tε1x1 . . . tεnxn,

f0 = tζ1y1 . . . tζkyk are cyclically reduced in G, with n ≥ 1. If fj /∈ gH0 for every j =

1, . . . ,m, then there is a finite group Q and an epimorphism ψ : A → Q such that forthe corresponding epimorphism ψ : G → P , extending ψ, with ψ(t) = s (where P is thespecial HNN-extension given by (8.7)), all of the following are true:

• ψ(fj) /∈ ψ(g0)ψ(H) in P , for each j ∈ 1, . . . ,m;• the elements ψ(g0) = sε1x1 . . . s

εnxn and ψ(f0) = sζ1 y1 . . . sζk yk are cyclically re-

duced in P , where xi := ψ(xi), i = 1, . . . , n, yl := ψ(yl), l = 1, . . . , k.

34 ASHOT MINASYAN

Proof. For every j = 1, . . . ,m, since fj /∈ gH0 in G, we can apply Lemma 8.7 (as H ≤ Ais a special subgroup of G), to find a finite index normal subgroup Lj C A, such that

ψj(fj) /∈ ψj(g0)ψj(H) in Pj, where ψj : G→ Pj is the homomorphism (obtained accordingto Remark 7.4) extending the natural epimorphism ψj : A→ A/Lj, and Pj is the specialHNN-extension of A/Lj with respect to ψj(H).

Now, since H is separable in A (by Lemma 8.1 and Corollary 5.4), there is a finite indexnormal subgroup K C A such that xi /∈ HK whenever xi /∈ H, for all i = 1, . . . , n, andyl /∈ HK whenever yl /∈ H, for all l = 1, . . . , k. Define the finite index normal subgroup Lof A by L := L1 ∩ · · · ∩ Lm ∩K, and let ψ : A→ Q := A/L be the natural epimorphism.Observe that for each j, the map ψj factors through the map ψ. Hence, once we let

ψ : G → P be the extension of ψ as in the formulation of Lemma 8.8, the UniversalProperty of special HNN-extensions (Remark 7.4) will imply that ψj factors through ψ,

for every j = 1, . . . ,m. Consequently, ψ(fj) /∈ ψ(g0)ψ(H) in P , for each j ∈ 1, . . . ,m.The second assertion of Claim B holds due to the choice of K and because L ≤ K. ThusLemma 8.8 is proved.

Lemma 8.9. Let K CG be a normal subgroup of finite index, let B be a special subgroupof G with B ≤ A, and let an element g ∈ G \ A be represented by a reduced productx0t

ε1x1tε2 . . . tεnxn in G, with n ≥ 1. Then there is a finite group Q and an epimorphism

ψ : A → Q such that for the corresponding homomorphism ψ : G → P , extending ψ andobtained according to Remark 7.4, with ψ(t) = s (where P is the special HNN-extensiongiven by (8.7)), all of the following are true:

• Cψ(B)

(ψ(g)

)⊆ ψ (CB(g)K) in P ;

• ker(ψ) ≤ A ∩K and ker(ψ) ≤ K.

Proof. Since A is residually finite (Lemma 8.1), the special subgroup H = HH is closedin PT (A) by Corollary 5.4. Therefore there exists a finite index normal subgroup M1CAsuch that xi /∈ HM1 in A for all i = 1, . . . , n − 1. As usual, we can replace M1 withM1 ∩K1, to make sure that M1 ≤ K1, where K1 := A ∩K.

Note that, according to Lemma 7.12, CB(g) = I in G, where

I := CB(x) ∩ x0Hx−10 ∩ (x0x1)H(x0x1)−1 ∩ · · · ∩ (x0 . . . xn−1)H(x0 . . . xn−1)−1,

and x := x0 . . . xn ∈ A.

Since rank(A) = r − 1 < r, by the induction hypothesis the claims of Lemmas 8.5 and8.6 hold for A. Hence, we can use Lemma 8.3 to find a finite index normal subgroup L1CAsuch that L1 ≤ M1 ≤ K1 and, for the corresponding epimorphism ψ : A → Q := A/L1,we have

J := CB(x) ∩ x0Hx−10 ∩ (x0x1)H(x0x1)−1 ∩ · · · ∩ (x0 . . . xn−1)H(x0 . . . xn−1)−1 ⊆ ψ(IM1)

in Q, where B, x, H and xi denote the ψ-images of B, x, H and xi in Q, i = 0, . . . , n.

Let P be the special HNN-extension of Q given by (8.7), and let ψ : G → P be

the extension of ψ provided by Remark 7.4, with ψ(t) = s. Since xi /∈ HL1 in A for


i = 1, . . . , n− 1, the product x0sε1x1 . . . s

εnxn is reduced and represents the element ψ(g)

in P . Consequently, Lemma 7.12 tells us that Cψ(B)

(ψ(g)

)= J in P . And noting that

ψ(M1) ≤ ψ(K1) = ψ(K1) ≤ ψ(K), we achieve

Cψ(B)

(ψ(g)

)= J ⊆ ψ(I)ψ(M1) ⊆ ψ(I)ψ(K) = ψ (CB(g)K) in P.

Finally, observe that ker(ψ) = L1 ≤ K1 = A ∩K and ker(ψ) is the normal closure of

L1 in G (by Lemma 7.5). And since L1 ≤ K C G, we see that ker(ψ) ≤ K in G. ThusLemma 8.9 has been established.

Lemma 8.10. Let g0 = tε1x1 . . . tεnxn be a cyclically reduced element in G, with n ≥ 1.

Then there exists an epimorphism ψ from A onto a finite group Q such that for thecorresponding extension ψ : G → P from G onto the special HNN-extension P of Q(given by (8.7)), with ψ|A = ψ and ψ(t) = s, we have

ker(ψ) ≤ A ∩K, ker(ψ) ≤ K in G, and CP

(ψ(g0)

)⊆ ψ

(CG(g0)K

)in P.

Proof. Clearly there exists m ∈ 0, 1, . . . , n such that we can enumerate all the prefixesp1, . . . , pn+1 of g0 so that p−1

j g0pj /∈ gH0 in G whenever 1 ≤ j ≤ m, and p−1j g0pj ∈ gH0 in G

whenever m+ 1 ≤ j ≤ n+ 1. For each j ∈ m+ 1,m+ 2, . . . , n+ 1, choose hj ∈ H suchthat hjp

−1j g0pjh

−1j = g0 in G, and set Ω := hjp−1

j |m+ 1 ≤ j ≤ n+ 1 ⊂ G.

Let f0 := g0 = tε1x1 . . . tεnxn and fj := p−1

j g0pj for j = 1, . . . ,m. Applying Lemma 8.8to g0, f0, . . . , fm ∈ G we find a finite group Q1, an epimorphism ψ1 : A → Q1, thespecial HNN-extension P1 of Q1 with respect to ψ1(H), and the corresponding extension

ψ1 : G → P1 of ψ1 (obtained by Remark 7.4), such that ψ1(fj) /∈ ψ1(g0)ψ1(H) in P1, for

each j ∈ 1, . . . ,m, and the element ψ1(g0) is cyclically reduced in P1.

On the other hand, by Lemma 8.9, there exist a finite group Q2, an epimorphismψ2 : A → Q2, the special HNN-extension P2 of Q2 with respect to ψ2(H), and the

corresponding extension ψ2 : G → P2 of ψ2, such that Cψ2(H)

(ψ2(g0)

)⊆ ψ2 (CH(g0)K)

in P2, ker(ψ2) ≤ A ∩K and ker(ψ2) ≤ K.

Define a finite index normal subgroup L0 C A by L0 := ker(ψ1) ∩ ker(ψ2) ≤ A ∩ K,and let ψ : A → Q := A/L0 be the natural epimorphism. By Remark 7.4, there is an

epimorphism ψ : G → P , extending ψ so that ψ(t) = s, where P is the special HNN-extension of Q given by (8.7). Since ker(ψ) = L0 ≤ ker(ψi), the maps ψi : A→ Qi factorthrough ψ for i = 1, 2. Consequently, according to the Universal Property of specialHNN-extensions (Remark 7.4), the maps ψi : G → Pi factor through ψ : G → P fori = 1, 2. Therefore we have

(8.9) ψ(g0) is cyclically reduced and ψ(fj) /∈ ψ(g0)ψ(H) in P , ∀ j ∈ 1, . . . ,m.

On the other hand, since ker(ψ) ≤ ker(ψ2) ≤ K in G, we also have

ψ−1(Cψ(H)

(ψ(g0)

))⊆ ψ−1

2

(Cψ2(H)

(ψ2(g0)

))⊆ CH(g0)K in G,

36 ASHOT MINASYAN

which implies

(8.10) Cψ(H)

(ψ(g0)

)⊆ ψ (CH(g0)K) in P.

Case 1: xn ∈ H. Then, according to Proposition 7.8, n = 1, CG(g0) = 〈t〉CH(g0) in G,

and CP

(ψ(g0)

)= 〈s〉Cψ(H)

(ψ(g0)

)in P . Recalling (8.10), we see that

CP

(ψ(g0)

)⊆⟨ψ(t)

⟩ψ (CH(g0)K) = ψ (CG(g0)K) in P.

Case 2: xn ∈ A \H. In this case (8.9) implies that ψ(pm+1), . . . , ψ(pn+1) is the list of

all prefixes of ψ(g0) satisfying ψ(pj)−1ψ(g0)ψ(pj) ∈ ψ(g0)ψ(H), because if 1 ≤ j ≤ m, then

ψ(pj)−1ψ(g0)ψ(pj) = ψ(p−1

j g0pj) = ψ(fj) /∈ ψ(g0)ψ(H) in P .

Therefore, by Proposition 7.8, CP

(ψ(g0)

)= Cψ(H)

(ψ(g0)

)⟨ψ(g0)

⟩Ω, where Ω :=

ψ(hj)ψ(pj)−1 |m+ 1 ≤ j ≤ n+ 1 = ψ(Ω) ⊂ P . Thus, recalling (8.10), we achieve

CP

(ψ(g0)

)⊆ ψ

(CH(g0)K〈g0〉Ω

)= ψ(CG(g0)K) in P.

In either of the two cases we have shown that CP

(ψ(g0)

)⊆ ψ(CG(g0)K) in P . This

completes the proof of Lemma 8.10.

We are finally ready to prove the two main Lemmas 8.5 and 8.6 announced above.

Proof of Lemma 8.5. There are two separate cases to consider.

Case 1: B 6= G.

Choose a maximal special subgroup A of G containing B. Then A is a right angled Artingroup of rank r − 1, and, according to Remark 7.3, G splits as a special HNN-extension(8.6) of A with respect to some special subgroup H of A.

If g ∈ A, then gB is closed in PT (A) by the induction hypothesis. Since G is residuallyfinite (Lemma 8.1), gB is separable in G by Lemma 5.5.

Thus we can suppose that g ∈ G \ A. Take any element f ∈ G \ gB. Let Q, P ,

ψ : A→ Q and ψ : G→ P be given by Lemma 8.7, so that ψ(f) /∈ ψ(g)ψ(B) in P .

Observe that P is a virtually free group as an HNN-extension of the finite group Q,

hence P is residually finite. Since ψ(B) = ψ(B) ⊆ Q is finite, ψ(g)ψ(B) is a finite subsetof P . Hence there is a homomorphism ξ : P → R from P to a finite group R such that

ξ(ψ(f)

)/∈ ξ

(ψ(g)ψ(B)

)in R. Consequently, the homomorphism ϕ : G → R, defined

by ϕ := ξ ψ, satisfies the condition ϕ(f) /∈ ϕ(gB). Therefore we have shown that gB isseparable in G in Case 1.

Case 2: B = G.

If g = 1 then gG = 1 is separable in G because G is residually finite (Lemma 8.1).Hence we can suppose that g 6= 1. But then, by Lemma 6.8, there exists a maximal


special subgroup A of G such that g /∈ AG. The group G is a special HNN-extension (8.6)of A with respect to a certain special subgroup H ≤ A (by Remark 7.3). Obviously, g isconjugate in G to some cyclically reduced element g0 = tε1x1 . . . t

εnxn with n ≥ 1, becauseg /∈ AG. This implies that gG = gG0 in G.

To show that gG is closed in PT (G), consider any element f ∈ G \ gG.

Sub-case 2.1: suppose, at first, that f /∈ AG. Then we can find a cyclically reducedelement f0 = tζ1y1 . . . t

ζmym ∈ fG. Let f1, f2, . . . , fm be the list of all cyclic permutationsof f0 in G.

Observe that fj /∈ gH0 ⊂ gG for every j = 1, . . . ,m, because f0 /∈ gG. Therefore we can

apply Lemma 8.8 to find Q, P , ψ : A→ Q and ψ : G→ P from its claim.

Since ψ(f1), . . . , ψ(fm) is the list of all cyclic permutations of ψ(f0) in P , Lemma

8.8, together with Lemma 7.7, imply that ψ(f0) /∈ ψ(g0)P in P . Now, according toLemma 8.2, there is a homomorphism ξ : P → R such that R is a finite group and

ξ(ψ(f0)

)/∈ ξ(ψ(g0)

)R. Therefore, defining the homomorphism ϕ : G→ R by ϕ := ξ ψ

we achieve ϕ(f0) /∈ ϕ(g0)R in R. But since ϕ(f) is conjugate to ϕ(f0) and ϕ(g0) isconjugate to ϕ(g) in R, we can conclude that ϕ(f) /∈ ϕ(g)R = ϕ(gG) in R.

To finish proving Case 2, it remains to consider

Sub-case 2.2: f ∈ AG. Set m := 0 and denote f0 = g0 ∈ G. Applying Lemma 8.8 to g0

and f0, we can find a homomorphism ψ from G to a special HNN -extension P of a finitegroup Q such that ψ(g0) = sε1ψ(x1) . . . sεnψ(xn) is cyclically reduced in P . Since n ≥ 1,

by Lemma 7.7 we have ψ(g0) /∈ ψ(A)P = ψ(AG) in P , hence ψ(f) /∈ ψ(g0)P = ψ(g)P inP . Arguing as above, we can find a finite quotient R of P (and, hence, of G) such thatthe images of f and g are not conjugate in R.

We can now conclude that the conjugacy class gG is closed in PT (G). Thus Case 2 isconsidered. This finishes the proof of Lemma 8.5.

Proof of Lemma 8.6. Take any element g ∈ G and any finite index normal subgroupK CG. As in Lemma 8.5, the proof splits into two main cases.

Case 1: B 6= G.

Choose a maximal special subgroup A of G containing B. Then A is a right angledArtin group of rank r−1 < r, and G is the special HNN-extension (8.6) of A with respectto a certain special subgroup H ≤ A (by Remark 7.3). Define the finite index normalsubgroup K1 of A by K1 := K ∩ A.

Sub-case 1.1: g ∈ A. Then, according to the induction hypothesis, the pair (B, g)satisfies the Centralizer Condition CCA in A, hence there exists L1 C A such that |A :L1| <∞, L1 ≤ K1, and the natural epimorphism ψ : A→ Q := A/L1 satisfies

(8.11) Cψ(B)(ψ(g)) ⊆ ψ (CB(g)K1) in Q.

Let ρA : G → A be the canonical retraction and set L := ρ−1A (L1) ∩K. Then L C G,

|G : L| < ∞, L ≤ K and ρA(L) = L1 ≤ K1 (since K1 = K ∩ A ⊆ ρA(K)). Let

38 ASHOT MINASYAN

ϕ : G→ R := G/L be the natural epimorphism. Observe that ker(ψ) = ker(ϕ) ∩A in G.Indeed, ker(ψ) = L1, ker(ϕ) = L, and L1 ⊆ ρ−1

A (L1)∩K∩A = L∩A, L∩A ⊆ ρA(L) = L1.

Therefore, without loss of generality, we can assume that Q ≤ R, and the restriction ofϕ to A coincides with ψ. Then we have ψ(K1) = ϕ(K1) ⊆ ϕ(K) in R. Since g ∈ A andB ≤ A, (8.11) implies that

Cϕ(B)(ϕ(g)) = Cψ(B)(ψ(g)) ⊆ ψ (CB(g))ψ(K1) ⊆ ϕ (CB(g))ϕ(K) in R,

which shows that the pair (B, g) has CCG in Sub-case 1.1.

Sub-case 1.2: g ∈ G \ A. Then the element g can be represented as a reduced productx0t

ε1x1tε2 . . . tεnxn in G, with n ≥ 1. Therefore we can find the groups Q, P and the maps

ψ : A → Q, ψ : G → P from the claim Lemma 8.9, so that all of the assertions of thatlemma hold.

Note that the subgroup ψ(B)∩ ψ(K) ≤ Q ≤ P is finite, therefore, since P is residually

finite (as a virtually free group), the finite set ψ(g)ψ(B)∩ψ(K) is separable in P . Conse-quently, by Lemma 3.7, there exists a finite group R and an epimorphism ξ : P → R suchthat ker(ξ) ≤ ψ(K) and

Cξ(ψ(B))

(ξ(ψ(g)

))⊆ ξ

(Cψ(B)

(ψ(g)

)ψ(K)

)in R.

Define the epimorphism ϕ : G → R by ϕ := ξ ψ, and observe that ker(ϕ) =

ψ−1 (ker(ξ)) ⊆ ψ−1(ψ(K)

)= K ker(ψ). But ker(ψ) ≤ K according to the second asser-

tion of Lemma 8.9, hence L := ker(ϕ) ≤ K in G.

Finally, recalling the first assertion of Lemma 8.9, we see that

Cϕ(B) (ϕ(g)) = Cξ(ψ(B))

(ξ(ψ(g)

))⊆ ξ

(Cψ(B)

(ψ(g)

)ψ(K)

)⊆ ξ

(ψ (CB(g)K) ψ(K)

)= ϕ(CB(g)K) in R.

Thus we have shown that the pair (B, g) has CCG in Sub-case 1.2. Therefore, B has CCG

in Case 1.

Case 2: B = G.

The pair (G, 1) evidently satisfies CCG, therefore we can assume that g 6= 1. In thiscase, by Lemma 6.8, there exists a maximal special subgroup A of G such that g /∈ AG.By Remark 7.3, G splits as a special HNN-extension (8.6) of A with respect to a certainspecial subgroup H ≤ A. Obviously, there exists z ∈ G such that g = zg0z

−1 in G, forsome cyclically reduced element g0 = tε1x1 . . . t

εnxn, where n ≥ 1 because g /∈ AG.

Now we apply Lemma 8.10 to find Q, P , ψ : A → Q and ψ : G → P from its claim,

so that ker(ψ) ≤ K and CP

(ψ(g0)

)⊆ ψ(CG(g0)K) in P . Note that P is virtually free

(being an HNN-extension of a finite group Q), hence every subgroup of P is virtually freeas well. Therefore, by Lemma 8.2, P is hereditarily conjugacy separable, and, thus, byProposition 3.2, P satisfies the Centralizer Condition CC.


Consequently, there exists a finite group R and an epimorphism ξ : P → R such thatker(ξ) ≤ ψ(K) and

CR

(ξ(ψ(g0)

))⊆ ξ

(CP

(ψ(g0)

)ψ(K)

)in R.

Defining the epimorphism ϕ : G → R by ϕ := ξ ψ, and arguing in the same manneras in Sub-case 1.2, we can show that L := ker(ϕ) ≤ K and CR (ϕ(g0)) ⊆ ϕ(CG(g0)K) inR. Conjugating both sides of the latter inclusion by ϕ(z) (and recalling that g = zg0z

−1

in G), we achieve CR (ϕ(g)) ⊆ ϕ(CG(g)K).

Hence B = G has CCG in Case 2, and Lemma 8.6 is proved.

Thus Lemmas 8.5 and 8.6 have been proved when rank(G) = r. Therefore, by inductionthey are true for all r ∈ N∪0, and we are ready to prove the main result of this paper.

Proof of Theorem 1.1. Let G be a right angled Artin group associated to a finite simplicialgraph Γ. Then for every g ∈ G, the conjugacy class gG is separable in G by Lemma 8.5.And Lemma 8.6 tells us that G satisfies the Centralizer Condition CC. Therefore, byProposition 3.2, G is hereditarily conjugacy separable.

9. Applications to separability properties

The first two applications that we mention do not directly follow from the statementof Theorem 1.1, but are consequences of its proof.

Corollary 9.1. Let A and B be conjugates of special subgroups of a right angled Artingroup G. Then for any element x ∈ G, the double coset AxB is separable in G.

Proof. Evidently, it is enough to consider the case when A and B are special subgroups ofG. Then A and B are retracts of G and the corresponding retractions commute (Remark6.1). By Remark 6.2, A ∩ B is also a special subgroup of G, hence Lemma 8.5 impliesthat the subset αA∩B is separable in G for every α ∈ G. Therefore, AxB is separable inG by Lemma 5.3.

In the case when x = 1 and A,B are special subgroups of G (not conjugates of them),Corollary 9.1 was proved by Haglund and Wise in [34, Cor. 9.4] using different arguments,based on Niblo’s criterion for separability of double cosets (see [46]). Unfortunately, ingeneral this criterion cannot be used to prove separability of double cosets of the formAxB if x ∈ G is an arbitrary element (because the retractions onto A and xBx−1 may nolonger commute).

Similarly, using Lemmas 8.5 and 8.6 together with Lemmas 6.5 and 5.7, we can obtainthe following:

Corollary 9.2. Suppose that A and D are conjugates of special subgroups in a rightangled Artin group G, and g ∈ G is an arbitrary element. Then the double coset CA(g)Dis separable in G.

40 ASHOT MINASYAN

The rest of applications in this section discuss conjugacy separability of various groups.Let us start with the following well-known observation:

Lemma 9.3. If H is a retract of a conjugacy separable group G, then H is conjugacyseparable.

Proof. Indeed, let ρH ∈ End(G) be a retraction of G onto H. Suppose that x, y ∈ Hand y /∈ xH in H. If there existed g ∈ G such that y = gxg−1 in G, then we wouldhave y = ρH(y) = ρH(g)ρH(x)ρH(g)−1 = ρH(g)xρH(g)−1 in H, contradicting to ourassumption. Therefore, y /∈ xG, and since G is conjugacy separable, there is a finitegroup R and a homomorphism ϕ : G → R such that ϕ(y) /∈ ϕ(x)R. Let Q := ϕ(H) andψ : H → Q ≤ R be the restriction of ϕ to H. By construction, we have that ψ(y) /∈ ψ(x)Q

in Q. Therefore H is conjugacy separable.

Remark 9.4. If F is a finite index subgroup in a virtual retract H of a group G, then Fitself is a virtual retract of G.

Indeed, let K ≤ G be a finite index subgroup containing H, and let ρH be a retractionof K onto H. Then M := ρ−1

H (F ) ≤ K has finite index in K, and, hence, in G. Evidentlythe restriction of ρH to M is a retraction of M onto F .

Combining Remark 9.4 with Lemma 9.3 we obtain the following statement (cf. [11,Thm. 3.4]):

Lemma 9.5. A virtual retract of a hereditarily conjugacy separable group is hereditarilyconjugacy separable itself.

Next comes a classical fact about conjugacy separable groups:

Lemma 9.6. Suppose G is a group satisfying the Unique Root property. If G contains aconjugacy separable subgroup H of finite index, then G is conjugacy separable.

Proof. Consider any element x ∈ G. We need to show that the conjugacy class xG isseparable in G.

Assume, first, that x ∈ H. Then xH is closed in PT (H), and since |G : H| < ∞,

it is also closed in PT (G). Choose z1, . . . , zk ∈ G so that G =⊔ki=1 ziH. Then xG =⋃k

i=1 zixHz−1

i is a finite union of closed sets in PT (G), hence xG is separable in G.

Now, if x ∈ G is an arbitrary element, then there is n ∈ N such that g := xn ∈ H,and, as we have shown above, gG is separable in G. Take any y ∈ G \ xG. Since G hasthe Unique Root property, we see that yn /∈ gG. Hence, there exists a finite index normalsubgroup NCG such that yn /∈ gGN in G. Consequently, y /∈ xGN (because the inclusiony ∈ xGN implies the inclusion yn ∈ (xn)GN). Thus G is conjugacy separable.

It is easy to see that Lemma 9.6 can be generalized as follows:

Lemma 9.7. If a group G has the Unique Root property and contains a hereditarily con-jugacy separable subgroup of finite index, then G is itself hereditarily conjugacy separable.


Note that the assumption of Lemma 9.6 demanding G to satisfy the Unique Rootproperty is important: in [27] A. Goryaga constructed an example of a finitely generatedgroup G which is not conjugacy separable, but contains a conjugacy separable subgroupof index 2.

Corollary 9.8. If a group G ∈ AVR has the Unique Root property, then G is hereditarilyconjugacy separable.

Proof. Let K ≤ G be a subgroup of finite index. By definition, G contains a finiteindex subgroup H which is a virtual retract of some right angled Artin group A. Since|H : (K ∩H)| ≤ |G : K| ≤ ∞, K ∩H is a virtual retract of A by Remark 9.4. But theindex |K : (K ∩H)| is also finite, hence K ∈ AVR.

Now, Theorem 1.1 and Lemma 9.5 imply that K ∩H is conjugacy separable. Hence Kis conjugacy separable by Lemma 9.6. Thus G is hereditarily conjugacy separable.

Recall that two groups G1 and G2 are said to be commensurable, if there exist finiteindex subgroups H1 ≤ G1 and H2 ≤ G2 such that H1 is isomorphic to H2. The proof ofCorollary 9.8 allows to conclude that the class AVR is closed under passing to subgroupsof finite index. Therefore we can make

Remark 9.9. If G1 is commensurable to G2 and G1 ∈ AVR, then G2 ∈ AVR.

As we observed in Lemma 6.3, right angled Artin groups have the Unique Root property.Another well-known class of groups with this property is the class of torsion-free wordhyperbolic groups.

Lemma 9.10. Torsion-free word hyperbolic groups have the Unique Root property.

Proof. Let G be a torsion-free word hyperbolic group. Suppose that xn = yn in G forsome x, y ∈ G and n ∈ N. Since G is torsion-free, we can assume that the orders of x andy are infinite. It is well-known that every element g ∈ G, of infinite order, belongs to aunique maximal virtually cyclic subgroup E(g) ≤ G (see, for instance, [48, Lemma 1.16]).

Note that the element g := xn ∈ G has infinite order and g ∈ E(x) ∩ E(y). Therefore,E(x) = E(y), thus y ∈ E(x). But since G is torsion-free, the virtually cyclic subgroupE(x) ≤ G must be cyclic. That is, there exists z ∈ G such that x = zk and y = zl forsome k, l ∈ Z. Obviously, the equality xn = yn implies that k = l. Thus x = y, and,hence, G enjoys the Unique Root property.

Combining Lemma 9.10 with Corollary 9.8 we obtain

Corollary 9.11. If G ∈ AVR is a torsion-free word hyperbolic group, then G is heredi-tarily conjugacy separable.

Using Osin’s results from [49], it is not difficult to generalize Lemma 9.10 as follows:if a group G is torsion-free and hyperbolic relative to a collection of proper subgroups,each of which has the Unique Root property, then G has the Unique Root property. Asa result, Corollary 9.11 can also be restated for this kind of relatively hyperbolic groups.

42 ASHOT MINASYAN

We can also establish Corollary 2.4, mentioned in Section 2.

Proof of Corollary 2.4. Since H has finite index in G, it is also word hyperbolic ([29]),thus, according to Lemma 9.10, H enjoys the Unique Root property.

By Corollary 2.3, G has a hereditarily conjugacy separable subgroup F ≤ G of finiteindex. Define K := H∩F ≤ G. Then K will be hereditarily conjugacy separable (because|F : K| < ∞). And since |H : K| < ∞, Lemma 9.7 implies that H is hereditarilyconjugacy separable.

10. Applications to outer automorphism groups

We have already discussed a few applications of Theorem 1.1 to outer automorphismgroups in Section 2. This section’s aim is to prove Theorem 2.6.

We refer the reader to Osin’s monograph [50] for the definition and basic properties ofrelatively hyperbolic groups. All relatively hyperbolic groups that we consider here arehyperbolic relative to families of proper subgroups. In the sense of B. Farb [25], this wouldcorrespond to weak relative hyperbolicity together with the Bounded Coset PenetrationCondition (the equivalence of Osin’s and Farb’s definitions for finitely generated groupsis proved in [50, Thm. 6.10]).

The following lemma is not difficult to prove but its statement is very useful (see, forexample, [31, Lemma 5.4]).

Lemma 10.1. Suppose that G is a finitely generated group and N is a centerless normalsubgroup of finite index in G. Then some finite index subgroup of Out(G) is isomorphic toa quotient of a subgroup of Out(N) by a finite normal subgroup. In particular, if Out(N)is residually finite, then Out(G) is residually finite.

Recall, that a group G is called elementary, if it contains a cyclic subgroup of finiteindex.

Lemma 10.2. If G is a non-elementary relatively hyperbolic group, then its center Z(G)is finite.

Proof. Suppose that G is hyperbolic relative to a family of proper non-trivial subgroupsHλλ∈Λ.

First, if |Λ| = ∞, then G splits as a non-trivial free product by [50, Thm. 2.44], and,thus, Z(G) = 1. If the set Λ is finite and each parabolic subgroup Hλ, λ ∈ Λ, isfinite, then G is word hyperbolic (in the sense of Gromov) by [50, Cor. 2.41]. And it iswell-known that the center of a non-elementary word hyperbolic group is finite.

Therefore we can assume that there is some µ ∈ Λ such that |Hµ| =∞. A theorem ofOsin [50, Thm. 1.4] claims that the intersection Hµ∩gHµg

−1 is finite for every g ∈ G\Hµ.If z ∈ Z(G), then Hµ = Hµ ∩ zHµz

−1 is infinite, hence z ∈ Hµ, i.e., Z(G) ⊆ Hµ.On the other hand, there exists g ∈ G \ Hµ because Hµ is a proper subgroup of G.By Osin’s theorem, Hµ ∩ gHµg

−1 is finite. And since Z(G) ⊆ gHµg−1, we see that

Z(G) ⊆ Hµ ∩ gHµg−1 must be finite.


The proof of Theorem 2.6 will use the following fact, established in [43, Cor. 1.4]:

Lemma 10.3. If G is a torsion-free non-elementary relatively hyperbolic group, thenAutpi(G) = Inn(G).

Proof of Theorem 2.6. Let G be a relatively hyperbolic group from the class AVR. IfG is virtually cyclic, then Out(G) is finite (cf. [43, Lemma 6.6]). Hence we can furthersuppose that G is non-elementary. By the assumptions, G contains a finite index subgroupN ∈ VR, and, in view of Remark 9.4, we can assume that N CG.

Note that N is finitely generated (and even finitely presented), as a virtual retract ofa finitely presented group. And since N has finite index in G, it is non-elementary andrelatively hyperbolic. The latter is an immediate consequence of Bowditch’s definition ofrelatively hyperbolic groups given in [6, Def. 2] (which is equivalent to Osin’s definition,as shown in [50, Thm. 6.10]); this also follows from the powerful result of C. Drutu [19,Thm. 1.2], which claims that relative hyperbolicity is invariant under quasi-isometries.

By construction, N is a virtual retract of some right angled Artin group A. And sinceA is torsion-free, N ≤ A is torsion-free as well. Therefore, according to Lemma 10.2,Z(N) = 1. The group N is finitely generated and conjugacy separable by Corollary2.1, and Autpi(N) = Inn(N) by Lemma 10.3. Hence we can apply Grossman’s theorem[30, Thm. 1] to conclude that Out(N) is residually finite. Consequently, Out(G) isresidually finite by Lemma 10.1.

11. Applications to the conjugacy problem

As it was shown by Mal’cev [38] and Mostowskii [44], a finitely presented conjugacyseparable group has solvable conjugacy problem. This result can be generalized as follows:

Lemma 11.1. Suppose that H is a finitely generated subgroup of a finitely presentedgroup G, such that for every h ∈ H the H-conjugacy class hH is separable in G. Thenthe conjugacy problem for H is solvable.

Proof. Without loss of generality we can assume that G = 〈X ‖R〉, for some finite set Xand a finite set of words R over the alphabet X±1, and H is generated by a finite subsetY of X. Let F (X) denote the free group on the set X and let F (Y ) be the subgroupof F (X) generated by Y . Then the identity map on X gives rise to the epimorphismθ : F (X)→ G, such that ker θ = N is the normal closure of R in F (X).

Since N is the normal closure of only finitely many words in F (X) and Y is finite, astandard argument (cf. [44]) shows that there is a partial algorithm A, which, given tworeduced words U,W ∈ F (Y ), terminates if and only if U ∈ W F (Y )N (i.e., if θ(U) ∈ θ(W )H

in G). The algorithm A lists every word from W F (Y )N in F (X), freely reduces it andcompares it with U ; it stops once it finds a word in W F (Y )N that is equal to U in F (X).

On the other hand, as G ∼= F (X)/N is finitely presented, there is an effective procedurelisting all homomorphisms ψ from F (X) to all finite groups Q, satisfying N ⊆ kerψ (see[44]). Given such a homomorphism ψ and any two reduced words U,W ∈ F (Y ), one

44 ASHOT MINASYAN

can decide in finitely many steps whether or not ψ(U) ∈ ψ(W )ψ(F (Y )) in Q, becauseψ(F (Y )) = 〈ψ(Y )〉 and Y is finite.

For any U,W ∈ F (Y ) denote u := θ(U) and w := θ(W ). If u /∈ wH , the separability ofwH in G implies the existence of a finite group Q and a homomorphism φ : G→ Q suchthat φ(u) /∈ φ(wH) = φ(w)φ(H) in Q. Thus the homomorphism ψ := φ θ : F (X) → Qsatisfies N ⊆ kerψ and ψ(U) /∈ ψ(W )ψ(F (Y )) in Q. And, of course, the existence of sucha homomorphism ψ tells us that u /∈ wH in G.

Hence, there is a partial algorithm B, which takes on input two words U,W ∈ F (Y )and terminates if and only if θ(U) ∈ θ(W )H in G. This algorithm goes through all thehomomorphisms ψ from F (X) to finite groups Q with N ⊆ kerψ, and stops when it findsone such that ψ(U) /∈ ψ(W )ψ(F (Y )) in Q.

The solution of the conjugacy problem for H amounts to taking on input two reducedwords U, V ∈ F (Y ) and running the two partial algorithms A and B simultaneously. One(and only one) of these two algorithms will eventually terminate, thus answering whetheror not θ(U) is conjugate to θ(W ) in H.

Corollary 11.2. Let G be a hereditarily conjugacy separable group. Suppose that H isa subgroup of G such that the double coset CG(h)H is separable in G for every h ∈ H.Then H is conjugacy separable. If, in addition, G is finitely presented and H is finitelygenerated, then H has solvable conjugacy problem.

Proof. The first claim is a direct consequence of Proposition 3.2 and Corollary 3.5. Theyalso imply that hH is separable in G for every h ∈ H. Therefore, the second claim followsfrom Lemma 11.1.

We are now in a position to prove Theorem 2.8, announced in Section 2.

Proof of Theorem 2.8. In [56] Servatius completely described centralizers of elements inright angled Artin groups. In particular, it follows from his description that CG(h) isfinitely generated for every h ∈ G.

Let ψ : G → G/N be the natural epimorphism and consider any h ∈ N . ThenE := ψ(CG(h)) is a finitely generated subgroup of G/N , hence E is closed in PT (G/N)by the assumptions. Since the map ψ is continuous (when G and G/N are consideredas topological groups equipped with their profinite topologies), we can conclude thatCG(h)N = ψ−1(E) is closed in PT (G) for every h ∈ N .

Therefore the claim of Theorem 2.8 follows from Theorem 1.1 and Corollary 11.2.

Corollary 11.3. Let N be a finitely generated normal subgroup of a right angled Artingroup G such that the quotient G/N is virtually polycyclic. Then every finite index sub-group K of N is conjugacy separable and has solvable conjugacy problem.

Proof. Since N is finitely generated, K contains a characteristic subgroup L of N with|N : L| < ∞. And since N C G, we can conclude that L C G, and the group G/L isan extension of the finite group N/L by the virtually polycyclic group G/N . An easy


induction on the length of the series with cyclic quotients shows that every finite-by-polycyclic group is polycyclic-by-finite. Thus G/L is virtually polycyclic, hence it issubgroup separable – see [55, Ex. 11 in Chapter 1.C].

Arguing as in the proof of Theorem 2.8, we see that the double coset CG(h)L is separable

in G for each h ∈ G. But K =⋃ki=1 Lxi for some k ∈ N and x1, . . . , xk ∈ K. Therefore

CG(h)K =⋃ki=1CG(h)Lxi is separable in G (as a finite union of separable subsets) for all

h ∈ K.

Note that K is finitely generated as a finite index subgroup of N , hence K is conjugacyseparable and has solvable conjugacy problem by Theorem 1.1 and Corollary 11.2.

12. Appendix: the Centralizer Condition in profinite terms

Our intention here is to prove that for residually finite groups the condition CC fromthe Definition 3.1 is equivalent to the condition (3.2) of Chagas and Zalesskii. We referthe reader to the book [54] for the background on profinite completions.

Proposition 12.1. Let H be a subgroup of a residually finite group G and let g ∈ G.The following are equivalent:

1) the pair (H, g) satisfies the condition CCG from Definition 3.3;

2) CH(g) = CH(g), where H ≤ G is the closure of H in the profinite completion Gof G.

Proof. The profinite completion G of G is the inverse limit of finite quotients of G. There

is a canonical embedding of G into the Cartesian product∏

N∈N G/N , where N is the

set of all finite index normal subgroups of G. Thus G can be equipped with the producttopology, making it a compact topological group (each finite group G/N is endowed withthe discrete topology).

For each N ∈ N let ψN denote the natural epimorphism from G to G/N . Then the

map ψ : G → G, defined by ψ(x) := (ψN(x))N∈N for every x ∈ G, is a homomorphism.

And since G is residually finite, ψ is injective. Therefore we can assume that G ≤ G, andso the condition 2) makes sense. Every homomorphism ψM , M ∈ N , can be uniquely

extended to a continuous homomorphism ψM : G→ G/M (ψM can also be regarded as a

restriction to G of the canonical projection from∏

N∈N G/N to G/M).

First, suppose that the pair (H, g) satisfies CCG. Consider any h ∈ H such that

h /∈ CH(g). Then there exists K ∈ N such that ψK(h) /∈ ψK(CH(g)). Hence, by CCG,there is L ∈ N satisfying L ≤ K and ψ−1

L

(CψL(H)(ψL(g))

)⊆ CH(g)K = ψ−1

K (ψK(CH(g))).

Therefore ψK factors through ψL, hence ψL(h) /∈ ψL(CH(g)K). Consequently, ψL(h) /∈CψL(H)(ψL(g)), and so h /∈ CH(g). Thus we established the inclusion CH(g) ⊆ CH(g).Since the inverse inclusion is evident, we have proved that 1) implies 2).

Now, let us assume that the condition 2) holds. Choose any K ∈ N and denoteL := L ∈ N | L ≤ K. Arguing by contradiction, suppose that for each L ∈ L there is

46 ASHOT MINASYAN

xL ∈ H such that ψL(xL) ∈ CψL(H)(ψL(g)) \ (ψL(CH(g)K)). Note that L is a directed set

(if L1, L2 ∈ L then L1 L2 if and only if L2 ⊆ L1), hence (xL)L∈L is a net in G. Since G

is compact, this net has a cluster point h ∈ H ≤ G.

Consider any N ∈ N and set L = N ∩ K ∈ L. Then, according to the definition

of the topology on G, there is M ∈ L such that M ⊆ L and ψL(xM) = ψL(h). By

construction, ψL(xM) ∈ CψL(H)(ψL(g)), hence ψL(h) ∈ CψL(H)(ψL(g)), implying that

ψN(h) ∈ CψN (H)(ψN(g)) because L ≤ N . Since the latter holds for every N ∈ N , we canconclude that h ∈ CH(g).

On the other hand, since h is a cluster point of the net (xL)L∈L and K ∈ L, there exists

M ∈ L such that ψK(xM) = ψK(h). But since M ≤ K we have xM /∈ CH(g)KM =

CH(g)K = ψ−1K (ψK(CH(g))). Thus ψK(h) = ψK(xM) /∈ ψK(CH(g)), which implies that

h /∈ CH(g).

Thus we found an element h ∈ CH(g) \ CH(g), contradicting to the condition 2).Consequently, 2) implies 1).

Proposition 12.1 implies that for residually finite groups the Centralizer Condition CCfrom Definition 3.1 is equivalent to the condition (3.2) introduced by Chagas and Zalesskiiin [11]:

Corollary 12.2. A is residually finite group G satisfies CC if and only if CG(g) = CG(g)for every g ∈ G.

It is well known that conjugacy separability of a residually finite group G is equivalentto the condition

(12.1) gG ∩G = gG in G, for all g ∈ G.

In other words, the condition (12.1) says that two elements g and g′ of G are conjugate

in G if and only if they are conjugate in G.

We are now able to reformulate the hereditary conjugacy separability of G in purelyprofinite terms:

Corollary 12.3. Suppose that G is a residually finite group. Then G is hereditarilyconjugacy separable if and only if for every g ∈ G both of the following hold in the

profinite completion G of G:

• gG ∩G = gG;• CG(g) = CG(g).

Proof. The necessity follows from Proposition 3.2 and Corollary 12.2.

The sufficiency is given by the result of Chagas and Zalesskii [11, Prop. 3.1]. It canalso be deduced by first applying Corollary 12.2 and then Proposition 3.2.


References

[1] A. Abrams, Configuration Spaces and Braid Groups of Graphs. PhD thesis, University of Californiaat Berkeley (2000).

[2] J.M. Alonso, M.R. Bridson, Semihyperbolic groups. Proc. London Math. Soc. (3) 70 (1995), no. 1,56–114.

[3] P. Bahls, Automorphisms of Coxeter groups. Trans. Amer. Math. Soc. 358 (2006), no. 4, 1781–1796.[4] G. Baumslag, Automorphism groups of residually finite groups. J. London Math. Soc. 38 (1963),

117–118.[5] M. Bestvina, N. Brady, Morse theory and finiteness properties of groups. Invent. Math. 129 (1997),

no. 3, 445–470.[6] B. Bowditch, Relatively hyperbolic groups. Preprint, 1999.

Available at http://www.warwick.ac.uk/ masgak/preprints.html

[7] M.R. Bridson, On the subgroups of semihyperbolic groups. Essays on geometry and related topics,Vol. 1, 2, 85–111, Monogr. Enseign. Math., 38, Enseignement Math., Geneva, 2001.

[8] M.R.Bridson, C.F. Miller III, Structure and finiteness properties of subdirect products of groups.Proc. Lond. Math. Soc. (3) 98 (2009), no. 3, 631–651.

[9] I. Bumagin, D.T. Wise, Every group is an outer automorphism group of a finitely generated group.J. Pure Appl. Algebra 200 (2005), no. 1-2, 137–147.

[10] P.-E. Caprace, B. Muhlherr, Reflection triangles in Coxeter groups and biautomaticity. J. GroupTheory 8 (2005), no. 4, 467–489.

[11] S.C. Chagas, P.A. Zalesskii, Finite index subgroups of conjugacy separable groups. Forum Mathe-maticum 21 (2009), no. 2, 347–353.

[12] S.C. Chagas, P.A. Zalesskii, Limit groups are conjugacy separable. Internat. J. Algebra Comput. 17(2007), no. 4, 851–857.

[13] R. Charney, An introduction to right-angled Artin groups. Geom. Dedicata 125 (2007), 141–158.[14] R. Charney, K. Vogtmann, Finiteness properties of automorphism groups of right-angled Artin

groups. Bull. Lond. Math. Soc. 41 (2009), no. 1, 94–102.[15] E. Chesebro, J. Deblois, H. Wilton, Some virtually special hyperbolic 3-manifold groups. Preprint,

2009. arXiv:0903.5288v1[16] J. Crisp, E. Godelle, B. Wiest, The conjugacy problem in subgroups of right-angled Artin groups. J.

Topol. 2 (2009), no. 3, 442–460.[17] J. Crisp, B. Wiest, Embeddings of graph braid and surface groups in right-angled Artin groups and

braid groups. Algebr. Geom. Topol. 4 (2004), 439–472.[18] M.B. Day, Peak reduction and finite presentations for automorphism groups of right-angled Artin

groups. Geom. Topol. 13 (2009), no. 2, 817–855.[19] C. Drutu, Relatively hyperbolic groups: geometry and quasi-isometric invariance. Comm. Math. Helv.

84 (2009), No. 3, pp. 503-546.[20] G. Duchamp, D. Krob, The lower central series of the free partially commutative group. Semigroup

Forum 45 (1992), no. 3, 385–394.[21] G. Duchamp, J.-Y. Thibon, Simple orderings for free partially commutative groups. Internat. J.

Algebra Comput. 2 (1992), no. 3, 351–355.[22] A.J. Duncan, I.V. Kazachkov, V.N. Remeslennikov, Parabolic and quasiparabolic subgroups of free

partially commutative groups. J. Algebra 318 (2007), no. 2, 918–932.[23] J.L. Dyer, Separating conjugates in free-by-finite groups. J. London Math. Soc. (2) 20 (1979), no. 2,

215–221.[24] E.S. Esyp, I.V. Kazatchkov, V.N. Remeslennikov, Divisibility theory and complexity of algorithms for

free partially commutative groups. Groups, languages, algorithms, 319–348, Contemp. Math., 378,Amer. Math. Soc., Providence, RI, 2005.

[25] B. Farb, Relatively hyperbolic groups. Geom. Funct. Anal. 8 (1998), 810–840.[26] E. Formanek, Conjugacy separability in polycyclic groups. J. Algebra 42 (1976) 1–10.

48 ASHOT MINASYAN

[27] A.V. Goryaga, Example of a finite extension of an FAC-group that is not an FAC-group (Russian).Sibirsk. Mat. Zh. 27 (1986), no. 3, 203–205, 225.

[28] E.R. Green, Graph Products of Groups. PhD thesis,University of Leeds, 1990.[29] M. Gromov, Hyperbolic groups. Essays in group theory, 75–263, Math. Sci. Res. Inst. Publ., 8,

Springer, New York, 1987.[30] E.K. Grossman, On the residual finiteness of certain mapping class groups. J. London Math. Soc.

(2) 9 (1974/75), 160–164.[31] V. Guirardel, G. Levitt, The outer space of a free product. Proc. London Math. Soc. 94 (2007), no.

3, 695–714.[32] M. Gutierrez, A. Piggott, K. Ruane, On the automorphisms of a graph product of abelian groups.

Preprint, 2007. arXiv:0710.2573[33] F. Haglund, D.T. Wise, Coxeter groups are virtually special. Adv. Math. 224 (2010), no. 5, 1890–1903.[34] F. Haglund, D.T. Wise, Special cube complexes. Geom. Funct. Anal. 17 (2008), no. 5, 1551–1620.[35] S.P. Humphries, On representations of Artin groups and the Tits conjecture. J. Algebra 169 (1994),

no. 3, 847–862.[36] M.R. Laurence, A generating set for the automorphism group of a graph group. J. London Math.

Soc. (2) 52 (1995), no. 2, 318–334.[37] R.C. Lyndon, P.E. Schupp, Combinatorial group theory. Ergebnisse der Mathematik und ihrer Gren-

zgebiete, Band 89. Springer-Verlag, Berlin-New York, 1977. xiv+339 pp.[38] A.I. Mal’cev, On Homomorphisms onto finite groups (Russian). Uchen. Zap. Ivanovskogo Gos. Ped.

Inst. 18 (1958), 49–60.[39] A. Martino, A proof that all Seifert 3-manifold groups and all virtual surface groups are conjugacy

separable. J. Algebra 313 (2007), no. 2, 773–781.[40] A. Martino, A. Minasyan, Conjugacy in normal subgroups of hyperbolic groups. Preprint, 2009.

arXiv:0906.1606

[41] J. Meier, L. VanWyk, The Bieri-Neumann-Strebel invariants for graph groups. Proc. London Math.Soc. (3) 71 (1995), no. 2, 263–280.

[42] C. F. Miller III, On group theoretic decision problems and their classification. Annals of Math. Study68, Princeton University Press, Princeton, NJ, 1971.

[43] A. Minasyan, D. Osin, Normal automorphisms of relatively hyperbolic groups. Trans. Amer. Math.Soc. 362 (2010), 6079-6103.

[44] A.W. Mostowski, On the decidability of some problems in special classes of groups. Fund. Math. 59(1966), 123–135.

[45] B.H. Neumann, Groups covered by permutable subsets. J. London Math. Soc. 29 (1954), 236–248.[46] G.A. Niblo, Separability properties of free groups and surface groups. J. Pure Appl. Algebra 78 (1992),

no. 1, 77–84.[47] G.A. Niblo, L.D. Reeves, Coxeter groups act on CAT(0) cube complexes. J. Group Theory 6 (2003),

no. 3, 399–413.[48] A.Yu. Olshanskii,On residualing homomorphisms and G-subgroups of hyperbolic groups. Internat. J.

Algebra Comput. 3 (1993), no. 4, 365–409.[49] D.V. Osin, Elementary subgroups of relatively hyperbolic groups and bounded generation. Internat.

J. Algebra Comput. 16 (2006), no. 1, 99–118.[50] D.V. Osin, Relatively hyperbolic groups: intrinsic geometry, algebraic properties, and algorithmic

problems. Mem. Amer. Math. Soc. 179 (2006), no. 843.[51] V.N. Remeslennikov, Conjugacy in polycyclic groups (Russian). Algebra i Logika 8 (1969), 712–725.[52] V.N. Remeslennikov, Finite approximability of groups with respect to conjugacy (Russian). Sibirsk.

Mat. Z. 12 (1971), 1085–1099.[53] L. Ribes, D. Segal, P.A. Zalesskii, Conjugacy separability and free products of groups with cyclic

amalgamation. J. London Math. Soc. (2) 57 (1998), no. 3, 609–628.[54] L. Ribes, P. Zalesskii, Profinite groups. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge.

A Series of Modern Surveys in Mathematics, 40. Springer-Verlag, Berlin, 2000. xiv+435 pp.


[55] D. Segal, Polycyclic groups. Cambridge Tracts in Mathematics, 82. Cambridge University Press,Cambridge, 1983. xiv+289 pp.

[56] H. Servatius, Automorphisms of graph groups. J. Algebra 126 (1989), no. 1, 34–60.[57] P.F. Stebe, A residual property of certain groups. Proc. Amer. Math. Soc. 26 (1970), 37–42.[58] D.T. Wise, The residual finiteness of negatively curved polygons of finite groups. Invent. Math. 149

(2002), no. 3, 579–617.

(Ashot Minasyan) School of Mathematics, University of Southampton, Highfield, Sout-hampton, SO17 1BJ, United Kingdom.

E-mail address: [email protected]

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